All categories

2 years ago

Harold walks 3/4 mile in each 5/6 hour. Calculate Harold's unit rate. Explain how you found your answer.

1 answer:

6 0

To solve this problem, you must apply the proccedure shown below:

1. The problem gives the following information: <span>Harold walks 3/4 miles in each 5/6 hours. Therefore, you have to divide 3/4 miles by 5/6 hours, as below:

unit rate=(3/4)(5/6)
unit rate=(3x6)(4x5)
unit rate=18/20

3. When you simplify, you obtain:

unit rate=9/10

4. Therefore, as you can see, the answer is: Harold's unit rate is 9/10.</span>

You might be interested in

Your tutor asks you to record the time you spend using IT during the week. You have recorded this time below: Day Time Monday 0.

NikAS [45]

Let us convert all figures into decimals so that we can compare them easily.

Monday 0.3

Tuesday 15% = 0.15

Wednesday 1/6 = 0.1666

Thursday 0.2

Friday 1/8 = 0.125

Clearly, I spent the least amount of time on Friday using IT and the time is 0.125 or 1/8.

7 0

2 years ago

The coordinates of the points below represent the vertices of a rectangle.

andrey2020 [161]

Answer:

18 units

Step-by-step explanation:

5+5=10

4+4=8

8+10=18 units

5 0

2 years ago

Read 2 more answers

Ilene who is not yet 21 years old, is two years older than ida

vagabundo [1.1K]

Answer:

whats the question you need answered?

Step-by-step explanation:

0 0

2 years ago

A florist has 8 roses and 24 daffodils.she wants to make as nany identical bouquets as possible, with the same combination of ro

Paladinen [302]

They should have 3 daffodils and 1 rose in each bouquet and that will equal 8 bouquets.

8 0

1 year ago

A sample of 1714 cultures from individuals in Florida diagnosed with a strep infection was tested for resistance to the antibiot

Damm [24]

Answer:

a) $p$ represent the real population proportion of people who showed partial or complete resistance to the antibiotic

b) $\hat p=\frac{973}{1714}=0.568$ represent the estimated proportion of people who showed partial or complete resistance to the antibiotic

c) We are confident (95%) that the true proportion of individuals in Florida diagnosed with a strep infection was tested for resistance to the antibiotic penicillin present partial or complete resistance to the antibiotic is between 54.5% to 59.1%.

Step-by-step explanation:

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

Part a

Description in words of the parameter p

$p$ represent the real population proportion of people who showed partial or complete resistance to the antibiotic

$\hat p$ represent the estimated proportion of people who showed partial or complete resistance to the antibiotic

n=1714 is the sample size required

$z_{\alpha/2}$ represent the critical value for the margin of error

The population proportion have the following distribution

$p \sim N(p,\sqrt{\frac{p(1-p)}{n}})$

Part b

Numerical estimate for p

In order to estimate a proportion we use this formula:

$\hat p =\frac{X}{n}$ where X represent the number of people with a characteristic and n the total sample size selected.

$\hat p=\frac{973}{1714}=0.568$ represent the estimated proportion of people who showed partial or complete resistance to the antibiotic

Part c

The confidence interval for a proportion is given by this formula

$\hat p \pm z_{\alpha/2} \sqrt{\frac{\hat p(1-\hat p)}{n}}$

For the 95% confidence interval the value of $\alpha=1-0.95=0.05$ and $\alpha/2=0.025$ , with that value we can find the quantile required for the interval in the normal standard distribution.

$z_{\alpha/2}=1.96$

And replacing into the confidence interval formula we got:

$0.568 - 1.96 \sqrt{\frac{0.568(1-0.568)}{1714}}=0.545$

$0.568 + 1.96 \sqrt{\frac{0.56(1-0.56)}{1714}}=0.591$

And the 95% confidence interval would be given (0.545;0.591).

We are confident that about 54.5% to 59.1% of individuals in Florida diagnosed with a strep infection was tested for resistance to the antibiotic penicillin present partial or complete resistance to the antibiotic.

5 0

2 years ago