PLEASE HELP! YOU GET 30 POINTS IF YOU HELP!!
Which inequality is the solution to 20 + 2 < x/3 - 8
A) x > 10
B) x > 42
C) x > 90
D) x > 114

NO LINKS ANSWER PLSSS!!!

Answer:

A' (-2,-4) B' (-3,-7) C' (-1,-8) D' (1,-6)

Step-by-step explanation:

Just take it one point at a time and move each point down 3 and 4 left. Prime notation just means to put this little sign ' so it indicates that this is the new coordinate. So A was (2,-1) and A' (the new one) is (-2,-4)

Answer:

-6.78 x 10^-21

Step-by-step explanation:

-6.77626358E-21

= -6.78 x 10^-21

Answer:

37

Step-by-step explanation:

Answer:

land

Step-by-step explanation:

because,bus are more comfortable and easier other transportation are dangerous than land transportation ❤️

hope i helped u

Answer: option D

Step-by-step explanation: just did it

C was wrong and showed option D was the correct answer

Answer:

It's D: 2/5

Step-by-step explanation:

Parallel lines have the same slope.

Answer: #1: 15/12 aka 1 3/12 #2: < #3 11/9 aka 9 2/11

Step-by-step explanation: Hope it helps ya

Answer:

A.

Step-by-step explanation:

took test

Answer:

4

Step-by-step explanation:

Use PEMDAS. Use a calulator to solve 3 and 1/3 times 1.9+19.5 divided by 4 and 1/2 and after you solve that solve 62/75- 0.16. Divided the first answer by the second. Now move to 3.5+4 and 2/3+ 2 and 2/15 then solve. Next solve 0.5 times ( 1 and 1/20+ 4.1) and with the answer you got from the equation above divided that by the answer just solved. Now divided the number on the left with the number on the right and you should get 4.

Answer:

1

Step-by-step explanation:

1x1 is 1 because 1x1 is 1

Answer:

1

Step-by-step explanation:

Answer:

[tex](a)[/tex] [tex]\frac{2}{cos\theta}/\frac{1}{sin\theta} + \frac{4sin\theta}{cos\theta} =6tan\theta[/tex]

[tex](b)[/tex] [tex]\frac{2sin\theta}{cos\theta} + \frac{4sin\theta}{cos\theta} =6tan\theta[/tex]

(c) Addition and Quotient identity

Step-by-step explanation:

Given

[tex]\frac{2sec\theta}{csc\theta} + \frac{4sin\theta}{cos\theta} =6tan\theta[/tex] --- The expression missing from the question

Solving (a): Write the left hand side in terms of sin and cosine

In trigonometry:

[tex]sec\theta = \frac{1}{cos\theta}[/tex]

and

[tex]csc\theta = \frac{1}{sin\theta}[/tex]

So, the expression becomes:

[tex]\frac{2 * \frac{1}{cos\theta}}{\frac{1}{sin\theta}} + \frac{4sin\theta}{cos\theta} =6tan\theta[/tex]

[tex]\frac{\frac{2}{cos\theta}}{\frac{1}{sin\theta}} + \frac{4sin\theta}{cos\theta} =6tan\theta[/tex]

Rewrite as:

[tex]\frac{2}{cos\theta}/\frac{1}{sin\theta} + \frac{4sin\theta}{cos\theta} =6tan\theta[/tex]

Solving (b): Simplify

[tex]\frac{2}{cos\theta}/\frac{1}{sin\theta} + \frac{4sin\theta}{cos\theta} =6tan\theta[/tex]

Change / to *

[tex]\frac{2}{cos\theta}*\frac{sin\theta}{1} + \frac{4sin\theta}{cos\theta} =6tan\theta[/tex]

[tex]\frac{2sin\theta}{cos\theta} + \frac{4sin\theta}{cos\theta} =6tan\theta[/tex]

Solving (c): The property used

To do this, we need to further simplify

[tex]\frac{2sin\theta}{cos\theta} + \frac{4sin\theta}{cos\theta} =6tan\theta[/tex]

Take LCM

[tex]\frac{2sin\theta+ 4sin\theta}{cos\theta} =6tan\theta[/tex]

Add the numerator

[tex]\frac{6sin\theta}{cos\theta} =6tan\theta[/tex]

Apply quotient identity

[tex]\frac{sin\theta}{cos\theta} = tan\theta[/tex]

This gives

[tex]\frac{6sin\theta}{cos\theta} = 6tan\theta[/tex]

Hence, the properties applied are:

Addition and Quotient identity

Answer:

6.

Just count the x's on the graph...

Answer:

B. 4.73 and 4.97

Step-by-step explanation:

To solve this question, we need to understand the Empirical Rule and the Central Limit Theorem.

Empirical Rule:

The Empirical Rule states that, for a normally distributed random variable:

Approximately 68% of the measures are within 1 standard deviation of the mean.

Approximately 95% of the measures are within 2 standard deviations of the mean.

Approximately 99.7% of the measures are within 3 standard deviations of the mean.

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]s = \frac{\sigma}{\sqrt{n}}[/tex].

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]s = \frac{\sigma}{\sqrt{n}}[/tex].

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

Standard deviation of 0.92, sample of 500:

This means that [tex]\sigma = 0.92, n = 500, s = \frac{0.92}{\sqrt{500}} = 0.04[/tex]

By the central limit theorem, which interval do 99.7% of the sample means fall within?

Within 3 standard deviations of the mean. So

4.85 - 3*0.04 = 4.85 - 0.12 = 4.73

4.85 + 3*0.04 = 4.85 + 0.12 = 4.97

So, option B.

Answer:

4.73 and 4.97

Please give me brainliest, I really need it.

Answer:

D

Step-by-step explanation:

D goes thru positive 4 and negative 3

Answer:i dont know :(

Step-by-step explanation:

im sorry :(

Answer:

1. true

2. false

3. false

4. false

Step-by-step explanation:

hope this helps

please mark brainiest

please give 5 stars

thank you

Answer:

Step-by-step explanation:

[tex]T_j \sim exp ( \lambda j) \ \ \ j = 1 (1) n \\ \\ f_{Tj}(tj) = \lambda j e^{-\lambda j tj } , tj>0 \\ \\ P\Big[ Tj> x\Big] = \int \limits ^{\infty}_{x} \lambda j e^{-\lambda j tj}\ dtj \\ \\ = e^{-\lambda j x}, x > 0 \\ \\ \\ \\ a) F_x (x) \\ \\ = P[X \le x ] \\ \\ = 1 - P[X> x] \\ \\ = 1 - \pi \limits ^{n}_{j =1} \Big\{ P[T_j > x ] \Big \}[/tex]

[tex]\text{This is because the appliance has the capacity to work for more than (x) }[/tex][tex]\text{hours if and only if all the "n" components work more than (x) hours.}[/tex]

[tex]\text{Then:}[/tex]

[tex]= 1 - e \ \pi^{n}_{j=1} \ e^{-\lambda j x} \\ \\ = 1 - e^{- (\sum \limits ^{n}_{j=1} \lambda j)x}[/tex]

∴

[tex]CDF = 1 - e^{- (\sum \limits ^{n}_{j=1} \lambda j)x}\ , \ x>0[/tex]

[tex]PDF =\Big( \sum \limits ^{n}_{j =1} \lambda j\Big) e^{- (\sum \limits ^{n}_{j=1} \lambda j)x}\ , \ x>0[/tex]

[tex]f_x(x) = \Big( \sum \limits ^{n}_{j =1} \lambda j\Big) e^{- (\sum \limits ^{n}_{j=1} \lambda j)x}\ , \ x>0[/tex]

(b)

[tex]E(x) = \int \limits ^{\infty}_{o }x f_x (x) \ dx \\ \\ = \int \limits ^{\infty}_{o }x \ \Big( \sum \limits ^{n}_{j =1} \lambda j\Big) e^{- (\sum \limits ^{n}_{j=1} \lambda j)x}\ , \ dx[/tex]

[tex]= \dfrac{1}{\sum \limits ^n_{j=1} \lambda j} \ \ \int \limits ^{\infty}_{o} \Big [(\sum \limits ^n_{j=1} \lambda j )x \Big] e ^{-\Big ( \sum \limits ^{n}_{j=1} \lambda j \Big)x} \ \ d \Big( x \sum \limits ^n_{j=1} \lambda j \Big)[/tex]

[tex]= \dfrac{1}{\sum \limits ^n_{j= 1} \lambda j} \int \llimits ^{\infty}_{o} t e^{-t} \ dt[/tex]

[tex]= \dfrac{1}{\sum \limits ^n_{j= 1} \lambda j} \int \llimits ^{\infty}_{o} t e^{-t} \ dt \ \ \ \ \text{\Big[By transformation }t =( \sum \limits ^n_{j=1} \lambda j )x \Big][/tex]

[tex]= \dfrac{1}{\sum \limits ^n_{j= 1} \lambda j}[/tex]

(c)

[tex]Let \ \ l_y_{\dfrac{1}{2}} \text{ be the median}[/tex]

∴

[tex]F(l_y_{\dfrac{1}{2}}) = \dfrac{1}{2} \\ \\ 1 - e ^{-\Big (\sum \limits ^{n}_{j=1} \lambda j \Big)} {l_y}_{\frac{1}{2}} = \dfrac{1}{2} \\ \\ \dfrac{1}{2} = e^{-\Big (\sum \limits ^{n}_{j=1} \lambda j \Big) } {l_y}_{\frac{1}{2}} \\ \\ - In 2 = {-\Big (\sum \limits ^{n}_{j=1} \lambda j \Big)} {l_y}_{\frac{1}{2}} \\ \\ \\ \\ \mathbf{ {l_y}_{\frac{1}{2}} = \dfrac{ In \ 2}{\sum \limits ^{n}_{j=1} \lambda j }}[/tex]

Answer:

You will add!

Step-by-step explanation:

Words words words

Answer:

R" (0,3)

S" (1,1)

T" (4,2)

Step-by-step explanation:

The reflection of the point (x, y) across the x-axis is the point (x, -y).

Translate right 4- add 4 to the x coordinate

R ( -4, -3 )    R' ( -4 , 3)    R" ( 0 , 3)

S ( -3, -1)      S' (-3 , 1)      S"  ( 1 , 1)

T ( 0 , -2)     T' (0, 2)       T" ( 4 , 2)

Answer:

1. Has been answered below

2. Degree of freedom is 4

Step-by-step explanation:

Row 1: 12+10+18 = 40

Row 2: 14+18+13 = 45

Row 3: 66+142+210 = 418

Grand total = 40+45+418 = 503

Column 1 : 12+14+66 = 92

Column 2: 10+18+142 = 170

Column 3: 18+13+210 = 241

Total = 92+170+241 = 503

Expected frequencies of diseased workers

92x40/503 = 7.316

170x40/503 = 13.52

241 x 40/503 = 19.165

Expected frequencies of sensitized workers

92x45/503 = 8.23

170x45/503 = 15.20

241x 45/503 = 21.56

2.

Degree of freedom

= (Number of row -1)(number of column - 1)

= 3-1 * 3-1

= 2x2

= 4

Answer:

[tex]S_{25}=555[/tex]

Step-by-step explanation:

Given that,

In an AP, the 6th term is 39 i.e.

[tex]a_6=a+5d\\\\39=a+5d\ ....(1)[/tex]

In the same AP, the 19th term is 7.8 i.e.

[tex]a_{19}=a+18d\\\\7.8=a+18d\ ......(2)[/tex]

Subtract equation (1) from (2).

7.8 - 39 = a+18d - (a+5d)

-31.2 = a +18d-a-5d

-31.2 = 13d

d = -2.4

Put the value of d in equation (!).

39 = a+5(-2.4)

39 = a- 12

a = 39+12

a = 51

The sum of n terms of an AP is given by :

[tex]S_n=\dfrac{n}{2}[2a+(n-1)d][/tex]

Put n = 25 and respected values,

[tex]S_{25}=\dfrac{25}{2}[2(51)+24(-2.4)]\\\\=555[/tex]

Hence, the sum of first 25 terms of the AP is equal to 555.

Answer:

$4,515.14

Step-by-step explanation:

First we need to get the amount after 25 years on $400,000 at 5% interest

Using the compound interest formula

A = P(1+r)^n

A = 400,000(1+0.05)^25

A = 400,000(1.05)^25

A = 400,000(3.3864)

A = $1,354,541.98

Since 25years = (25*12)months

25 years = 300months

Amount to be able to pull out each month = $1,354,541.98/300

Amount to be able to pull out each month = $4,515.14

Hence you will be able to pull out $4,515.14 each month

Answer:

We fail to reject the Null ;

(88.323, 130.677)

Step-by-step explanation:

Given the data :

117, 156, 89, 72, 116, 125, 101, 100

H0 : μ ≥ 100

H1: μ < 100

From the data: using calculator :

Mean, x = 109.5

Standard deviation, s = 25.33

The test statistic:

(x - μ) ÷ s/Sqrt(n)

(109.5 - 100) ÷ 25.33/sqrt(8)

Test statistic = 1.06

Since, same size is small, we use t test ;

Using the Pvalue calculator :

T at df = 8 - 1 = 7, α = 0.05

Pvalue = 0.161

Pvalue > α

0.161 > 0.05 ; Hence, we fail to reject H0.

B.)

C. I = mean ± Tcritical *s/√n

Tcritical = 2.365 at 95% and df = 7

109.5 ± 2.365(25.3264/√8)

109.5 ± 2.365(8.9542)

(109.5-21.1768 ; 109.5+21.1768)

(88.323, 130.677)

In conclusion true mean is in the confidence interval, so it is consistent.

Answer:

say 5 and 6

Step-by-step explanation

you just multiply it

Answer:

Step-by-step explanation:

The first step contains the error.

2x - 10 =    22            Add 10 to both sides

   + 10     +10

2x = 32                      Divide by 10

2x/2 = 32/2

x = 16

Answer is in the photo. I can't attach it here, but I uploaded it to a file hosting. link below! Good Luck!

tinyurl.com/wpazsebu

Step-by-step explanation:

y>3-8

y>-5

The solutions are in the interval ]-5,○○[