A. 1/20 <br> B.1/60<br> C. 1/12 <br> D. 1/15

The football team has a total of 350 jerseys there are 154 medium-sized jerseys. What percent of the jerseys are medium-sized j

Lena [83]

Answer:

44% of the jerseys are medium sized.

Step-by-step explanation:

Total jerseys football team has = 350

Medium-sized jerseys = 154

We need to find what percent of the jerseys are medium-sized jerseys.

The formula used will be: $Percent\:of\:medium\:size\:jerseys=\frac{Medium\:size\:jerseys}{Total\:jersey}\times 100$

Putting values and finding percentage

$Percent\:of\:medium\:size\:jerseys=\frac{Medium\:size\:jerseys}{Total\:jersey}\times 100$

$Percent\:of\:medium\:size\:jerseys=\frac{154}{350}\times 100\\Percent\:of\:medium\:size\:jerseys=0.44\times 100\\Percent\:of\:medium\:size\:jerseys=44\%$

So, 44% of the jerseys are medium sized.

6 0

2 years ago

Find the area and please show the steps and explain. Thanks!

Anit [1.1K]

I'm pretty sure you just add all the numbers up so it would be 569.9 ft

3 0

3 years ago

Round to nearest hundred 64,505

NARA [144]

64,500 would be the answer

7 0

3 years ago

Read 2 more answers

A study by Allstate Insurance Co. finds that 82% of teenagers have used cell phones while driving (the Wall Street Journal, May

Strike441 [17]

The probability that the sample proportion is within ± 0.02 of the population proportion is 0.3328

<h3>How to determine the probability?</h3>

The given parameters are:

Sample size, n = 100
Population proportion, p = 82%

Start by calculating the mean:

$\mu = np$

$\mu = 100 * 82\%$

$\mu = 82$

Calculate the standard deviation:

$\sigma = \sqrt{\mu(1 - p)$

$\sigma = \sqrt{82 * (1 - 82\%)$

$\sigma = 3.84$

Within ± 0.02 of the population proportion are:

$x_{min} = 82 * (1 - 0.02) = 80.38$

$x_{max} = 82 * (1 + 0.02) = 83.64$

Calculate the z-scores at these points using:

$z = \frac{x - \mu}{\sigma}$

So, we have:

$z_1 = \frac{80.36 - 82}{3.84} = -0.43$

$z_2 = \frac{83.64 - 82}{3.84} = 0.43$

The probability is then represented as:

P(x ± 0.02) = P(-0.43 < z < 0.43)

Using the z table of probabilities, we have:

P(x ± 0.02) = 0.3328

Hence, the probability that the sample proportion is within ± 0.02 of the population proportion is 0.3328

A. 1/20 B.1/60 C. 1/12 D. 1/15