Juan will spend at most on gifts. So far, he has spent . What are the possible additional amounts he will spend?

A farmer knows that a grocery store will reject a shipment of his vegetables if more than 4% of the vegetables contain blemishes

kykrilka [37]

Answer:

Assumption A is violated

Step-by-step explanation:

x = number of blemishes = 8

n = sample size = 150

proportion = x/n = 8/150 = 0.0533

1. large counts: np> 10

= 150 * 0.0533 >10

= 7.995 > 10

this assumpton is obviously violated. 7.995 is not greater than 10

2. Large Counts: n(1 - p) > 10

150(1-0.0533)>10

150-7.995 > 10

142.005> 10

there is no violation. The assumption is satisfied

3. This assumption is satisfied. this is because the tomatoes were selected using simple random sampling.

4. 10% of the population = 0.1 * 2000 = 200

the sample size = 150

150 < 200

this assumption is satisfied

3 0

2 years ago

What is 5 ÷ 1/5 I really need help with dividing whole numbers by fractions

salantis [7]

Answer:

25

Step-by-step explanation:

When dividing by fractions, you can multiply by the opposite reciprocal.

5 ÷ 1/5

5 x 5/1

25

4 0

3 years ago

Read 2 more answers

The average length of a female dolphin is about 135 inches what is the length in feet and inches

Law Incorporation [45]

Eleven feet and two.five inches

8 0

3 years ago

Read 2 more answers

27. A bag contains 2 red and 4 yellow apples. Brandon selects 2 apples at random. What

djyliett [7]

Answer:12/30

Step-by-step explanation:

Total apple =6

Yellow apples = 4

4/6

3/5

(4/6)*(3/5)= 12/30

4 0

2 years ago

Calculate the sample standard deviation and sample variance for the following frequency distribution of hourly wages for a sampl

lidiya [134]

Answer:

(a) The sample variance is 16.51

(a) The sample standard deviation is 4.06

Step-by-step explanation:

Given

$\begin{array}{cc}{Class} & {Frequency} & 8.26 - 10.00 & 20 &10.01-11.75 & 38 &11.76 - 13.50& 36 & 13.51-15.25 &25&15.26-17.00 &27 &\ \end{array}$

Solving (a); The sample variance.

First, calculate the class midpoints.

This is the mean of the intervals.

i.e.

$x_1 = \frac{8.26+10.00}{2} = \frac{18.26}{2} = 9.13$

$x_2 = \frac{10.01+11.75}{2} = \frac{21.76}{2} = 10.88$

$x_3 = \frac{11.76+13.50}{2} = \frac{25.26}{2} = 12.63$

$x_4 = \frac{13.51+15.25}{2} = \frac{28.76}{2} = 14.38$

$x_5 = \frac{15.26+17.00}{2} = \frac{32.26}{2} = 16.13$

So, the table becomes:

$\begin{array}{ccc}{Class} & {Frequency} & {x} & 8.26 - 10.00 & 20&9.13 &10.01-11.75 & 38 &10.88&11.76 - 13.50& 36 &12.63& 13.51-15.25 &25&14.38&15.26-17.00 &27 &16.13\ \end{array}$

Next, calculate the mean

$\bar x = \frac{\sum fx}{\sum f}$

$\bar x = \frac{20*9.13 + 38 * 10.88+36*12.63+25*14.38+27*16.13}{20+38+36+25+27}$

$\bar x = \frac{1845.73}{146}$

$\bar x = 12.64$

Next, the sample variance is:

$\sigma^2 = \frac{\sum f(x - \bar x)^2}{\sum f - 1}$

So, we have:

$\sigma^2 = \frac{20*(9.13-12.63)^2 + 38 * (10.88-12.63)^2 +...........+27 * (16.13 -12.63)^2}{20+38+36+25+27-1}$

$\sigma^2 = \frac{2393.6875}{145}$

$\sigma^2 = 16.51$

The sample standard deviation is:

$\sigma = \sqrt{\sigma^2}$

$\sigma = \sqrt{16.51}$

$\sigma = 4.06$

6 0

2 years ago