Kyle has these 3 cards.

each day last week I sold 32, 45, 47, 63 and 72 product accessories. What was the median number of accessories sold for the week

AnnZ [12381]

Answer:

The median value is 47.

Step-by-step explanation:

First, organize the numbers in ascending order.

The median's position is calculated as ((N+1)÷2)th term, where N is the amount of data.

= 6÷2 th term

= 3 rd term

4 0

4 months ago

On his first day of school, Kareem found the high temperature in degrees Fahrenheit to be 76.1°. He plans to use the function to

Inessa [12570]

The formula to convert Fahrenheit to Celsius is:

Celsius = (Fahrenheit - 32) x 5/9

C(76.1) = (76.1 - 32) x 5/9
= (44.1) x 5/9
= 24.5

Hence, 76.1°F is equal to 24.5°C.

5 0

3 months ago

Read 2 more answers

In a sample of 88 adults selected randomly from one town, it is found that 6 of them have been exposed to a particular strain of

tester [12383]

Response with clarification:

Let p denote the proportion of adults in the town who have encountered this flu strain.

According to the provided information

$H_0:p=0.08\\\\ H_a: p\neq0.08$

∵ $H_a$ this is a two-tailed test.

Test statistic:

$z=\dfrac{\hat{p}-p}{\sqrt{\dfrac{p(1-p)}{n}}}$

, where p= denotes the population proportion

${\hat{p}$ = signifies the sample proportion

n= represents the sample size

Setting n= 6 and ${\hat{p}=\dfrac{6}{88}\approx0.068$ and p=0.08

$z=\dfrac{0.068-0.08}{\sqrt{\dfrac{0.08(1-0.08)}{88}}}$

$z=\dfrac{-0.012}{0.0289199522192}\approx-0.415$

P-value for the two-tailed test:[2P(Z>|z|)

=2P(Z>|-0.415|)

=2P(Z>0.415) = 2[1-P(Z≤0.415)] [∵ P(Z>z)=1-P(Z≤z)]

=2(1-0.6609) [from the z-table]

=0.6782

Decision: Because the p-value(0.6782) exceeds the significance level of 0.01, we do not reject the null hypothesis.

This leads us to conclude that there is insufficient evidence to back the assertion that the percentage of all adults in this town exposed to this flu strain deviates from the national average of 8%.

8 0

2 months ago