The table shows three unique functions. (TABLE IN PIC) Which statements comparing the functions are true? Select three options. Only f(x) and h(x) have y-intercepts. Only f(x) and h(x) have x-intercepts. The minimum of h(x) is less than the other minimums. The range of h(x) has more values than the other ranges. The maximum of g(x) is greater than the other maximums.

Answer:

a) The student cannot receive an A in the class.

b) The student must score 119 in the third exams to make an A.  This is clearly not possible, since he cannot make 119 in a 100-points exam.

c) The student can make a B but he must score at least 84 in the third exam.

Step-by-step explanation:

To make an A, the student must score 315 (350 x 90%) in both home and the three exams.

The student who scored 35 (7 + 8 + 7 + 5 + 8) in the homework and 161 (81 + 80), getting a total of 196, is short by 119 (315 - 196) scores in making an A.

To make a B, the student must score 280 (350 x 80%) or higher but not reaching 315.

B ≥ 280 and < 315.

Since, the student had scored 196, he needs to score 84 and above to make a B in the last exam.

Answer:

x=-52

Step-by-step explanation:

1/3x=-17 1/3

x=-52

Answer:

  see attached

Step-by-step explanation:

The limit as x gets large is the ratio of the highest-degree terms. In most cases, the limit can be found by evaluating that ratio. Where an absolute value is involved, the absolute value of the highest-degree term is used.

If the ratio gives x to a positive power, the limit does not exist. If the ratio gives x to a negative power, the limit is zero.

The arrangement of functions according to the given condition

[tex]m(x)=\frac{4x^{2}-6 }{1-4x^{2} }[/tex]

[tex]h(x)=\frac{x^{3} -x^{2} +4}{1-3x^{2} }[/tex]

[tex]k(x)=\frac{5x+1000}{x^{2} }[/tex]

[tex]i(x)=\frac{x-1}{|1-4x| }[/tex]

[tex]g(x)=\frac{|4x-1|}{x-4}[/tex]

[tex]l(x)=\frac{5x^{2} -4}{x^{2} +1}[/tex]

[tex]f(x)=\frac{x^{2} -1000}{x-5}[/tex]

[tex]j(x)=\frac{x^{2}-1 }{|7x-1|}[/tex]

A limit is the value that  a function approaches as the input approaches some value.

According to the given question

[tex]l(x)=\frac{5x^{2} -4}{x^{2} +1}[/tex]

⇒[tex]\lim_{nx\to \infty} \frac{5x^{2} -1}{x^{2} +1}[/tex]

⇒[tex]\lim_{x \to \infty} \frac{x^{2} }{x^{2} } \frac{5-\frac{1}{x^{2} } }{1+\frac{1}{x^{2} } }[/tex]

= 5           ([tex]\frac{1}{x^{2} } = 0[/tex] ,as x tends to infinity  [tex]\frac{1}{x^{2} }[/tex] tends to 0)

[tex]i(x)=\frac{x-1}{|1-4x|}[/tex]

⇒[tex]\lim_{x \to \infty} \frac{x-1}{|1-4x|}[/tex] =  [tex]\lim_{x \to \infty} \frac{x}{x} \frac{1-\frac{1}{x} }{|\frac{-1}{4}+\frac{1}{x} | }[/tex]  =[tex]\frac{1}{\frac{1}{4} }[/tex] =[tex]\frac{1}{4}[/tex]

As x tends to infinity 1/x tends to 0, and |[tex]\frac{-1}{4}[/tex]| gives 1/4

[tex]f(x)= \frac{x^{2} -1000}{x--5}[/tex]

⇒[tex]\lim_{x \to \infty} \frac{x^{2} -1000}{x-5}[/tex]= [tex]\lim_{x \to \infty} \frac{x^{2} }{x} \frac{1-\frac{1000}{x^{2} } }{1-\frac{5}{x} }[/tex]= [tex]\lim_{x \to \infty} x\frac{1-\frac{1000}{x^{2} } }{1-\frac{5}{x} }[/tex] ⇒ limit doesn't exist.

[tex]m(x)=\frac{4x^{2}-6 }{1-4x^{2} }[/tex]

⇒[tex]\lim_{x\to \infty} \frac{4x^{2} -6}{1-4x^{2} }[/tex]=[tex]\lim_{x\to \infty} \frac{x^{2} }{x^{2} } \frac{4-\frac{6}{x^{2} } }{\frac{1}{x^{2} } -4}[/tex]  [tex]= \lim_{n \to \infty} \frac{4}{-4}[/tex] = -1

As x tends to infinity [tex]\frac{1}{x^{2} }[/tex] tends to 0.

[tex]g(x)=\frac{|4x-1|}{x-4}[/tex]

⇒[tex]\lim_{x\to \infty} \frac{|4x-1|}{x-4}[/tex] = [tex]\lim_{x \to \infty} \frac{|x|}{x} \frac{4-\frac{1}{x} }{1 -\frac{4}{x} } }[/tex] = 4

as x tends to infinity 1/x tends to 0

and |x|=x ⇒[tex]\frac{|x|}{x}=1[/tex]

[tex]h(x)=\frac{x^{3}-x^{2} +4 }{1-3x^{3} }[/tex][tex]\lim_{x \to \infty} \frac{x^{3} -x^{2} +4}{1-3x^{3} }[/tex][tex]= \lim_{x \to \infty} \frac{x^{3} }{x^{3} } \frac{1-\frac{1}{x}+\frac{4}{x^{3} } }{\frac{1}{x^{3} -3} }[/tex]  = [tex]\frac{1}{-3}[/tex] =[tex]-\frac{1}{3}[/tex]

[tex]k(x)=\frac{5x+1000}{x^{2} }[/tex]

[tex]\lim_{x \to \infty} \frac{5x+1000}{x^{2} }[/tex] = [tex]\lim_{x \to \infty} \frac{x}{x} \frac{5+\frac{1000}{x} }{x}[/tex] =0

As x tends to infinity 1/x tends to 0

[tex]j(x)= \frac{x^{2}-1 }{|7x-1|}[/tex]

[tex]\lim_{x \to \infty} \frac{x^{2}-1 }{|7x-1|}[/tex] = [tex]\lim_{x \to \infty} \frac{x}{|x|}\frac{x-\frac{1}{x} }{|7-\frac{1}{x}| }[/tex]  = [tex]\lim_{x \to \infty} 7x[/tex] = limit doesn't exist.

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Answer:

d = 3 in

Step-by-step explanation:

Since we are trying to find the diameter, and the diameter is given to us as 3 in, our diameter is 3 in.

Answer:

D. Two-sample chi-square

Step-by-step explanation:

A chi-square test is a test used to compare the data that is observed, from the data that is expected.

In a two-sample chi-square test the observed data should be similar to the expected data if the two data samples are from the same distribution.

The hypotheses of the two-sample chi-square test is given as:

H0: The two samples come from a common distribution.

Ha: The two samples do not come from a common distribution

Therefore, in this case, the best statistical test to utilize is the two-sample chi-square test.

Answer:

33.777 yd = 0.030886 km

Step-by-step explanation:

==>Given:

33.777 yd

==>Required:

Convert 33.777 yd to km to 6 decimal places, using the metric and U.S systems.

==>Solution:

To convert, note that 1 km = 1093.6133 yd.

Thus,

1 km = 1093.6133 yd

x km = 33.777 yd

Cross multiply

1 × 33.777 = 1093.6133 × x

33.777 = 1093.6133x

Divide both sides by 1093.6133, to solve for x

33.777/1093.6133 = x

0.03088569 = x

x ≈ 0.030886 (to 6 decimal places)

Therefore, 33.777 yd = 0.030886 km

Answer:

The second question

Step-by-step explanation:

The orca starts at -25 meters. She goes up 25 meters.

up 25 = +25

-25+25=0

Answer:

Option 2

Step-by-step explanation:

The orca swims at the elevation of -25 meters. The orca swims up 25 meters higher than before.

-25 + 25 = 0

Answer:

12x/2 or 52/2

Step-by-step explanation:

Ok, perimeter is length+length+width+width. 12x/2 and 52/2 could are probably the answers.

The greatest number is 6.23 times 10 superscript 12.

The number is written in the form [tex]a \times 10^b[/tex] where we have [tex]1 \leq a < 10[/tex]

The number b shows the order, which is the most important figure for which scientific notation is used. It tells us how much order large or small a value is in powers of 10. We can for a time, ignore the value of 'a' for two comparable quantities and only compare their orders(this type of comparison is useful when difference is too big, like size of human to size of a star etc sort of comparisons).

We are given that the number so;

A.6.23 x 10^12 is equivalent to rolling the decimal 12 times to the right.

B.6.23 x 10^8 is equivalent to rolling the decimal 8 times to the right.

C.6.23 x 10^-6 is equivalent to rolling the decimal 6 times to the left.

D.6.23 x 10^3 is equivalent to rolling the decimal 3 times to the right.

This shows the 10 has been multiplied by itself thrice.

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Answer:

2/10 for Rebecka and either 2/9 or 1/9 for Aaron depending on if Rebecka selects a poodle or not.

Step-by-step explanation:

do some math

The 4 angles need to add to 360.

2 of them are 70

The other two need to equal 360-140 = 220

They are both the same so one angle needs to equal 220/2 = 110

Now find x:

X + 60 = 110

Subtract 60 from both sides:

X = 50. The answer is D

Answer:

A 95% confidence interval for the mean number of chips in a bag of Chips Ahoy Cookies is [1187.96, 1288.44].

Step-by-step explanation:

We are given that statistics students at the Air Force Academy (no kidding) purchased some randomly selected bags of cookies and counted the chocolate chips.

Some of their data are given below; 1219, 1214, 1087, 1200, 1419, 1121, 1325, 1345, 1244, 1258, 1356, 1132, 1191, 1270, 1295, 1135.

Firstly, the pivotal quantity for finding the confidence interval for the population mean is given by;

                          P.Q.  =  [tex]\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }[/tex]  ~ [tex]t_n_-_1[/tex]

where, [tex]\bar X[/tex] = sample mean number of chocolate chips = [tex]\frac{\sum X}{n}[/tex] = 1238.2

            s = sample standard deviation = [tex]\sqrt{\frac{\sum (X-\bar X)^{2} }{n-1} }[/tex]  = 94.3

            n = sample of car drivers = 16

            [tex]\mu[/tex] = population mean number of chips in a bag

Here for constructing a 95% confidence interval we have used a One-sample t-test statistics because we don't know about population standard deviation.

So, 95% confidence interval for the population mean, [tex]\mu[/tex] is ;

P(-2.131 < [tex]t_1_5[/tex] < 2.131) = 0.95  {As the critical value of t at 15 degrees of

                                             freedom are -2.131 & 2.131 with P = 2.5%}  

P(-2.131 < [tex]\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }[/tex] < 2.131) = 0.95

P( [tex]-2.131 \times {\frac{s}{\sqrt{n} } }[/tex] < [tex]{\bar X-\mu}[/tex] < [tex]2.131 \times {\frac{s}{\sqrt{n} } }[/tex] ) = 0.95

P( [tex]\bar X-2.131 \times {\frac{s}{\sqrt{n} } }[/tex] < [tex]\mu[/tex] < [tex]\bar X+2.131 \times {\frac{s}{\sqrt{n} } }[/tex] ) = 0.95

95% confidence interval for [tex]\mu[/tex] = [ [tex]\bar X-2.131 \times {\frac{s}{\sqrt{n} } }[/tex] , [tex]\bar X+2.131 \times {\frac{s}{\sqrt{n} } }[/tex] ]

                                   = [ [tex]1238.2-2.131 \times {\frac{94.3}{\sqrt{16} } }[/tex] , [tex]1238.2+2.131 \times {\frac{94.3}{\sqrt{16} } }[/tex] ]

                                   = [1187.96, 1288.44]

Therefore, a 95% confidence interval for the mean number of chips in a bag of Chips Ahoy Cookies is [1187.96, 1288.44].

Answer:

aaaaha pues

Step-by-step explanation:

Answer:

what happened

Step-by-step explanation:

Answer:

ur pito is to small

Step-by-step explanation:

it to little

Answer:

$36,241.20

Step-by-step explanation:

Compounded Interest Rate Formula: A = P(1 + r/n)^nt

Since we are given P, r, n, and t, simply plug it into the formula:

A = 35000(1 + 0.035/4)⁴⁽¹⁾

A = 35000(1 + 0.00875)⁴

A = 35000(1.00875)⁴

A = 35000(1.03546)

A = 36241.2

Answer:

15

Step-by-step explanation:

PEMDAS

3x3 = 9

3+3 = 6

9+6 = 15

By the BODMAS rule we get, 3 + 3 × 3 + 3 =​ 15

The acronym BODMAS rule is used to keep track of the right sequence of operations to do when solving mathematical issues. Brackets (B), order of powers or roots (O), division (D), multiplication (M), addition (A), and subtraction (S) are all represented by this acronym (S).

3 + 3 × 3 + 3 =​

3 × 3 = 9

3 + 9 + 3 = 15.

Therefore, the correct answer is 15.

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Answer:

3(7) + 2* |7 - 8| - 12 = 11

Step-by-step explanation:

3(7) + 2* |7 - 8| - 12

21 + 2* |-1| - 12

21 + 2* 1 - 12

21 + 2 - 12

23 - 12 = 11

Hope this helps! :)

Answer:

y = 2x + 3

Step-by-step explanation:

Slope Formula: [tex]m=\frac{y_2-y_1}{x_2-x_1}[/tex]

Slope-Intercept Form: y = mx + b

Step 1: Find slope m

m = (-1 - 3)/(-2 - 0)

m = -4/-2

m = 2

y = 2x + b

Step 2: Rewrite equation

y = 2x + 3

*You are given y-intercept (0, 3), so simply add it to your equation.

Answer:

Step-by-step explanation:

5x(x + 3) + 6(x + 3)

The final answer is

( 5x + 6 ) ( x + 3 )

Hope this helps you

Answer:

Answer C

Step-by-step explanation:

You are balancing this equation out by subtracting 7x from both sides. This means you are using the subtraction property of equality.

Answer:

-3+(-5)

Checking our answer:

Adding this does indeed give -8

Answer:

Null hypothesis = H₀ = There food preferences among vole species are independent of one another.

Alternate hypothesis = H₁ = There is a relationship between voles and food preference.

Expected meadow vole/apple slices = 29.983051

Expected common vole/apple slices = 28.016949

Expected meadow vole/peanut butter-oatmeal = 31.016949

Expected common vole/peanut butter-oatmeal = 28.983051

Chi-square value = χ² = 2.154239

Degree of freedom = 1

Critical value = 3.841

χ² < Critical value

We failed to reject H₀

We do not have significant evidence at the given significance level to show that there is a relationship between voles and food preference.

Step-by-step explanation:

He frequently traps the moles and has noticed what appears to be a preference for a peanut butter-oatmeal mixture by the meadow voles vs apple slices are usually used in traps, where the common voles seem to prefer the apple slices.

So he conducted a study where he used a peanut butter-oatmeal mixture in half the traps and the normal apple slices in his remaining traps to see if there was a food preference between the two different voles.

Null hypothesis = H₀ = There food preferences among vole species are independent of one another.

Alternate hypothesis = H₁ = There is a relationship between voles and food preference.

Data collected by Dr. Pagels:

                                              meadow voles     common voles      Row Total

apple slices                                     26                          32                      58

peanut butter-oatmeal                   35                          25                     60

Column Total                                   61                          57                     118

Where 118 is the grand total.

The expected number is given by

Expected = (row total)×(column total)/grand total

Expected meadow vole/apple slices = 58×61/118

Expected meadow vole/apple slices = 29.983051

Expected common vole/apple slices = 58×57/118

Expected common vole/apple slices = 28.016949

Expected meadow vole/peanut butter-oatmeal = 60×61/118

Expected meadow vole/peanut butter-oatmeal = 31.016949

Expected common vole/peanut butter-oatmeal = 60×57/118

Expected common vole/peanut butter-oatmeal = 28.983051

The chi-square statistic value is given by

χ² = Σ(Observed - Expected)²/Expected

χ² = (26 - 29.983051)²/29.983051 + (32 - 28.016949)²/28.016949 + (35 - 31.016949)²/31.016949 + (25 - 28.983051)²/28.983051

χ² = 2.154239

The degrees of freedom is given by

DoF = (row - 1)×(col - 1)

For the given case, we have 2 rows and 2 columns

DoF = (2 - 1)×(2 - 1)

DoF = 1

The given level of significance = 0.05

The critical value from the chi-square table at α = 0.05 and DoF = 1 is found to be

Critical value = 3.841

Conclusion:

Reject H₀ If χ² > Critical value

We reject the Null hypothesis If the calculated chi-square value is more than the critical value.

For the given case,

χ² < Critical value

We failed to reject H₀

We do not have significant evidence at the given significance level to show that there is a relationship between voles and food preference.

Answer:

2. Precise but not accurate

Step-by-step explanation:

In a high precision, low accuracy case study, the measurements are all close to each other (high agreement between the measurements) but not near/or close to the center of the distribution (how close a measurement is to the correct value for that measurement)

u/4 - 4 = -20

Add 4 to both sides:

u/4 = -16

Multiply both sides by 4:

u = -64

Answer:

u=-64

Step-by-step explanation:

u/4 -4 = -20

First add 4 to both sides.

u/4=-16

Now multiply both sides by 4

u=-64

Answer:

48.3 cm²

Step-by-step explanation:

Let A be the area of the yellow region

A= the area of the square - the area of the quarter square

A= 15²-(15²*π)/4= 48.28≈ 48.3 cm²

Answer:

69.5309950592 cm²

Step-by-step explanation:

Area of Square:

Area = [tex]Length * Length[/tex]

Area = 18*18

Area = 324 square cm

Area of circle:

Diameter = 18 cm

Radius = 9 cm

Area = [tex]\pi r^2[/tex]

Area = (3.14)(9)²

Area = (3.14)(81)

Area = 254.469004941 square cm

Area of Shaded area:

=> Area of square - Area of circle

=> 324 - 254.469004941

=> 69.5309950592 cm²

Answer:

Subtract 2 and two-thirds from both sides of the equation

8 minus 2 and two-thirds = 5 and one-third

Substitute the value for r to check the solution.

Step-by-step explanation:

2 2/3  + r   = 8

Subtract 2 2/3 from each side

2 2/3  + r  - 2 2/3   = 8 - 2 2/3

r = 5 1/3

Check the solution

2 2/3 +5 1/3 =8

8 =8

Answer:

1, 3, 5

Step-by-step explanation:

edge

Answer:

The 99% confidence interval for the difference between the true average load for the fiberglass beams and that for the carbon beams is (-11.937, -6.863).

Step-by-step explanation:

We have to calculate a 99% confidence interval for the difference between the true average load for the fiberglass beams and that for the carbon beams.

The sample 1 (Fiberglass), of size n1=26 has a mean of 33.4 and a standard deviation of 2.2.

The sample 2 (Carbon), of size n2=26 has a mean of 42.8 and a standard deviation of 4.3.

The difference between sample means is Md=-9.4.

[tex]M_d=M_1-M_2=33.4-42.8=-9.4[/tex]

The estimated standard error of the difference between means is computed using the formula:

[tex]s_{M_d}=\sqrt{\dfrac{\sigma_1^2}{n_1}+\dfrac{\sigma_2^2}{n_2}}=\sqrt{\dfrac{2.2^2}{26}+\dfrac{4.3^2}{26}}\\\\\\s_{M_d}=\sqrt{0.186+0.711}=\sqrt{0.897}=0.9473[/tex]

The critical t-value for a 99% confidence interval is t=2.678.

The margin of error (MOE) can be calculated as:

[tex]MOE=t\cdot s_{M_d}=2.678 \cdot 0.9473=2.537[/tex]

Then, the lower and upper bounds of the confidence interval are:

[tex]LL=M_d-t \cdot s_{M_d} = -9.4-2.537=-11.937\\\\UL=M_d+t \cdot s_{M_d} = -9.4+2.537=-6.863[/tex]

The 99% confidence interval for the difference between the true average load for the fiberglass beams and that for the carbon beams is (-11.937, -6.863).

In this way, we can calculate the individual duration of each one and the duration time, knowing that the sample means:

The 99% confidence interval for the difference between the true average load for the fiberglass beams and that for the carbon beams is -11.937 and -6.863.

We have to calculate a 99% confidence interval for the difference between the true average load for the fiberglass beams and that for the carbon beams. The sample 1 (Fiberglass), of size n1=26 has a mean of 33.4 and a standard deviation of 2.2. The sample 2 (Carbon), of size n2=26 has a mean of 42.8 and a standard deviation of 4.3. The difference between sample means is Md=-9.4.

[tex]Sm_d= \sqrt{\frac{\sigma^2_1}{n_1} +\frac{\sigma^2_2}{n_2}} = \sqrt{(0.186)+(0.711) }= 0.9473[/tex]

The critical t-value for a 99% confidednce interval is t=2.678. The margin of error (MOE) can be calculated as:

[tex]MOE=t*8M_d = (2.678)(0.9473)= 2.537[/tex]

Then, the lower and upper bounds of the confidence interval are:

[tex]LL= M_d-t*SM_d = -9.4-2.537= -11.937\\UL= M_d+t*SM_d= -9.4+2.537= -6.863[/tex]

The 99% confidence interval for the difference between the true average load for the fiberglass beams and that for the carbon beams is (-11.937, -6.863).

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Answer:

Yup P is the right one having 62.26%

Step-by-step explanation:

Answer:

Yes

Step-by-step explanation:

We can set it up as

P = 33/53

Q = 20/48

R = 54/90

S = 44/83

This is because we are calculating the percent of yellow birds in the total amt. of birds in a specified park.

Now we calculate =>

P = 33/53 = around 0.62

Q = 20/48 = around 0.416

R = 54/90 = 0.6

S = 44/83 = around 0.53

We find that Park P has the greatest percentage and -->

Thus, Park P is our answer and yes, you are correct.

Answer:

  5

Step-by-step explanation:

Substitute the given values and do the arithmetic.

  [tex]0.3y+\dfrac{y}{z}=0.3\cdot 10+\dfrac{10}{5}=3+2=\boxed{5}[/tex]