Mathematics College

how many ft makes a height of 1.66m. ​

Answers

Answer 1

Answer:

5.44 meters

Step-by-step explanation:

We Know

0.3048 meter = 1 ft

How many ft makes a height of 1.66m?

We Take

1.66 ÷ 0.3048 ≈ 5.44 meters

So, the answer is 5.44 meters

Related Questions

Lily measured the lengths of 16 fish.
Use the graph below to estimate the lower and upper quartiles of the fish

Answers

Answer:

The lower quartile of the fish lengths is 15 inches

The upper quartile of the fish lengths is 30 inches

Step-by-step explanation:

(the cumulative frequency gives how much of the observations/measurements are under a certain value e.g, in our case, for lengths of 40 inches, all 16 of the fish have length in that range (less than or equal to 40) so, the cumulative frequency is 16 and so on)

We need to find the lower and upper quartiles of the fish lengths,

Now, the lower quartile contains 25% (or 1/4) of the values

in our case, since there are 16 fish,

25% of 16 is 4

So, for the lower quartile, the cumulative frequency should be 4.

For which the length is 15 inches,

So, the lower quartile of the fish lengths is 15 inches

The upper quartile contains 75% (or 3/4) of the values,

Now, 3/4 of 16 is 12.

So, for the upper quartile, the cumulative frequency should be 12.

For which the length is 30 inches

So, the upper quartile of the fish lengths is 30 inches

Determine the measure of the third angle in a triangle when the other two angles total 165 degrees.

Answers

Answer:

15

Step-by-step explanation:

180-15

Hello!

the sum of the angles in the triangle = 180°

so the 3rd angle = 180° - 165° = 15°

The answer is 15°

In a controlled laboratory experiment, scientists at the University of Minnesota discovered that
25% of a certain strain of rats subjected to a 20% coffee
bean diet and then force-fed a powerful cancer-causing
chemical later developed cancerous tumors. Would we
have reason to believe that the proportion of rats developing tumors when subjected to this diet has increased
if the experiment were repeated and 16 of 48 rats developed tumors? Use a 0.05 level of significance.

Answers

Yes, we would have reason to believe that the proportion of rats developing tumors when subjected to this diet has increased if the experiment were repeated and 16 of 48 rats developed tumors.

To determine whether there is an increase in the proportion of rats developing tumors when subjected to a coffee bean diet, we can conduct a hypothesis test using the 0.05 level of significance.

1. State the hypotheses:

  - Null hypothesis (H0): The proportion of rats developing tumors remains the same.

  - Alternative hypothesis (Ha): The proportion of rats developing tumors has increased.

2. Identify the test statistic:

  We will use a z-test to compare the observed proportion of rats developing tumors with the expected proportion.

3. Set the significance level:

  The significance level (α) is given as 0.05.

4. Collect data:

  In the original experiment, 25% of rats developed tumors. In the repeated experiment, 16 out of 48 rats developed tumors.

5. Compute the test statistic:

  The test statistic formula for comparing proportions is:

  z = (p - P) / sqrt(P(1-P)/n)

  where p is the observed proportion, P is the hypothesized proportion, and n is the sample size.

  Using the observed proportion (16/48 = 0.333), the hypothesized proportion (0.25), and the sample size (48), we can calculate the test statistic.

6. Determine the critical value:

  Since we are using a 0.05 level of significance and conducting a one-tailed test (Ha: >), we can find the critical value from the standard normal distribution table. The critical value for a 0.05 significance level is 1.645.

7. Make a decision:

  If the test statistic is greater than the critical value, we reject the null hypothesis and conclude that the proportion of rats developing tumors has increased.

8. Calculate the test statistic:

  Plugging in the values into the formula, we calculate the test statistic:

  z = (0.333 - 0.25) / sqrt(0.25 * 0.75 / 48) = 1.404

9. Compare the test statistic and critical value:

  The test statistic (1.404) is less than the critical value (1.645).

10. Make a decision:

   Since the test statistic is not greater than the critical value, we fail to reject the null hypothesis. Therefore, we do not have sufficient evidence to conclude that the proportion of rats developing tumors has increased when subjected to this diet.

In summary, based on the given data and conducting a hypothesis test, we do not have reason to believe that the proportion of rats developing tumors has increased if the experiment were repeated and 16 of 48 rats developed tumors.

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The July bank statement sent by the bank to ABC company shows a balance of cash on deposit at July 31 of Br.4,964.47 Assume that on July 31, assume the on July 31, ABCs ledger shows a bank balance of Br. 4 173. 83. 1. A deposit of Br 410. 90 made after banking hours and doesnt appear in the bank statement2. A check drawn for Br. 79 had been erroneously charged by the bank Br.973. For checks issued in July have not yet been paid by the bank (outstanding checks). : Theses checks are;- Check No date amount 801 June 15 ---Br.100,00 888 July 24 ----Br. 10.25 890 July 27--- Br. 294.50 891 July 30 ---Br. 205.004. A check written for birr 210 had been incorrectly charged by the bank as birr 120 5. Proceeds from collection of a interest bearing note receivable from David. ABC Company had left this note with the banks collection department. The face amount of the note was birr 500 6. Br. 24.75 interest earned on average account balance during July7. A check for Br. 10 returned with the statement had been recorded in the check register as Br. 100. The check was for the payment of an obligation to Davis Equipment Company for the purchase of office supplies on account 8. Br. 5,00 fee charged by bank for handling collection of note receivable 9. Br. 50.25 check from customer John deposited by ABC company charged bank as Non sufficient fund (NSF) 10. Br. 12.70 service charged by bank for the month of July. 11. Check number 875 was issued July 20 in the amount of Br 85 but was erroneously recorded in the cash payment Journal as Br 58 for payment of telephone expense

Answers

Based on the given information and adjustments, the adjusted bank balance for ABC Company on July 31 would be Br. 5,169.42.

Based on the provided information, let's analyze the various transactions and their impact on the bank reconciliation for ABC Company:

1. The deposit of Br. 410.90 made after banking hours: This deposit does not appear in the bank statement, so it should be added to the bank balance. After including this deposit, the adjusted bank balance would be Br. 5,375.37 (Br. 4,964.47 + Br. 410.90).

2. The erroneously charged check of Br. 79: This check should be deducted from the bank balance as it was charged in error by the bank. After deducting this amount, the adjusted bank balance would be Br. 5,296.37 (Br. 5,375.37 - Br. 79).

3. Outstanding checks: The outstanding checks that haven't been paid by the bank need to be deducted from the adjusted bank balance. The total amount of outstanding checks is Br. 610.75 (Br. 100 + Br. 10.25 + Br. 294.50 + Br. 205). After deducting this amount, the adjusted bank balance would be Br. 4,685.62 (Br. 5,296.37 - Br. 610.75).

4. Incorrectly charged check of Br. 210 as Br. 120: This error made by the bank needs to be corrected. The bank balance should be increased by the difference, which is Br. 90. Therefore, the adjusted bank balance would be Br. 4,775.62 (Br. 4,685.62 + Br. 90).

5. Proceeds from collection of an interest-bearing note receivable: The face amount of the note was Br. 500. Since the note was left with the bank's collection department, the amount collected should be added to the bank balance. Therefore, the adjusted bank balance would be Br. 5,275.62 (Br. 4,775.62 + Br. 500).

6. Interest earned on the average account balance: The interest earned of Br. 24.75 should be added to the bank balance. After including this amount, the adjusted bank balance would be Br. 5,300.37 (Br. 5,275.62 + Br. 24.75).

7. The returned check of Br. 10: This check was recorded in the check register as Br. 100 but was returned for Br. 10. The difference of Br. 90 should be deducted from the bank balance. Therefore, the adjusted bank balance would be Br. 5,210.37 (Br. 5,300.37 - Br. 90).

8. Fee charged by the bank for handling the collection of the note receivable: The fee of Br. 5.00 should be deducted from the bank balance. After deducting this fee, the adjusted bank balance would be Br. 5,205.37 (Br. 5,210.37 - Br. 5.00).

9. Check from customer John charged as Non-sufficient funds (NSF): The check of Br. 50.25 should be deducted from the bank balance. After deducting this amount, the adjusted bank balance would be Br. 5,155.12 (Br. 5,205.37 - Br. 50.25).

10. Service charge by the bank for the month of July: The service charge of Br. 12.70 should be deducted from the bank balance. After deducting this charge, the adjusted bank balance would be Br. 5,142.42 (Br. 5,155.12 - Br. 12.70).

11. Erroneously recorded check of Br. 85 as Br. 58: This error in the cash payment journal needs to be corrected. The bank balance should be increased by the difference, which is Br. 27. Therefore, the adjusted bank balance would be Br. 5,169.42 (Br. 5,142.42 + Br. 27).

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Nadia wants to find out if students in her year like the school lunches. She
decides to ask her class to complete a survey.
Which of the statements below describes:
ack to task
a) the population?
b) the sample?
All the students in the school
All the students in Nadia's class
All the students in Nadia's year
All the students not in Nadia's class
All the students not in Nadia's year

Answers

Answer:

a) All the students in Nadia's year are the population

b) All the students in Nadia's class are the sample

Step-by-step explanation:

She wants to find out if students in her year like the school lunches.

So, this is the population.

i.e All the students in Nadia's year

She asks her class to complete a survey,

So, this is the sample of the population (Her classmates are in the same year as her and so on)

i.e All the students in Nadia's class

Find the area of a composite figure.

Answers

The area of the composite figure is 800m²

What is area of a figure?

The area of a figure is the number of unit squares that cover the surface of a closed figure.

Composite geometric figures are made from two or more geometric figures.

The figure consist of a rectangle , a semi circle and a triangle.

Area of the semicircle = 1/2 πr²

= 1/2 × 3.14 × 10²

= 314/2 = 157 m²

Area of the rectangle = l × w

= 25 × 20

= 500m²

area of the triangle = 1/2bh

= 1/2 × 10 × 25

= 25 × 5

= 125 m²

Therefore the area of the composite figure

= 125 + 500 + 175

= 800m²

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