John and Alan have a collection of x baseball cards. John has x/4 cards. What fraction of the cards does Alan have?

In a survey conducted by a website, employers were asked if they had ever sent an employee home because they were dressed inappr

MA_775_DIABLO [31]

Answer:

Test statistic Z = p diff/std error = 2.3333

p value one tailed = 0.009815

Step-by-step explanation:

Given that in a survey conducted by a website, employers were asked if they had ever sent an employee home because they were dressed inappropriately.

Sample size n = 2755

Sample favourable x = 967

Sample proportion p = $\frac{967}{2755} \\=0.3510$

$H_0: p = 0.3333\\H_a: p >0.3333$

(right tailed test at 5% significance level)

p difference = 0.0210

Standard error assuming H0 is true is $\sqrt{\frac{0.3333*0.6667}{2755} } \\=0.0090$

Test statistic Z = p diff/std error = 2.3333

p value one tailed = 0.009815

Since p <0.05 we reject null hypothesis.

6 0

3 years ago

A field researcher is gathering data on the trunk diameters of mature pine and spruce trees in a certain area. The following are

lidiya [134]

Answer:

As the pvalue of the test is 0.02097 < 0.1, using a .10 significance level, he can conclude that the average trunk diameter of a pine tree is greater than the average diameter of a spruce tree.

Step-by-step explanation:

Using a .10 significance level, can he conclude that the average trunk diameter of a pine tree is greater than the average diameter of a spruce tree

The null hypothesis is that they are equal(that is, subtraction between the means is 0), while the alternate hypothesis is that they are greater(that is, subtraction between the means is larger than 0). So

$H_0: \mu_1 - \mu_2 = 0$

$H_a: \mu_1 - \mu_2 > 0$

In which $\mu_1$ is the mean pine trees height while $\mu_2$ is the mean spruce trees height.

The test statistic is:

$z = \frac{X - \mu}{s}$

In which X is the sample mean, $\mu$ is the value tested at the null hypothesis and s is the standard error.

0 is tested at the null hypothesis:

This means that $\mu = 0$

Mean trunk diameter (cm) 35 30:

Pine trees - 35

Spruce trees - 30

The sample mean is given by the subtraction of the means. So

$X = 35 - 30 = 5$

Sample Size 40 80

Population variance 160 160

This means that the standard error for each sample is given by:

$s_1 = \frac{\sqrt{160}}{\sqrt{40}} = 2$

$s_1 = \frac{\sqrt{160}}{\sqrt{80}} = \sqrt{2}$

The standard error of the difference is the square root of the sum squared of the standard error of each sample.

$s = \sqrt{2^2 + (\sqrt{2})^2} = \sqrt{6} = 2.4495$

Test statistic:

$z = \frac{X - \mu}{s}$

$z = \frac{5 - 0}{2.4495}$

$z = 2.04$

Pvalue of the test:

Probability of a sample mean larger than 5, which is 1 subtracted by the pvalue of Z when X = 5.

Looking at the z-table, z = 2.04 has a pvalue of 0.9793.

1 - 0.9793 = 0.0207

As the pvalue of the test is 0.02097 < 0.1, using a .10 significance level, he can conclude that the average trunk diameter of a pine tree is greater than the average diameter of a spruce tree.

8 0

3 years ago

18 tens 20 ones equal what hundreds

uysha [10]

200 is the answer so it would equal to 2 hundreds

8 0

3 years ago

HELPPPPPPPPPPPPPPPPPPPPPPPP

mixer [17]

Answer:

x = 20

Step-by-step explanation:

The angles shown are alternate exterior angles.

Alternate exterior angles are congruent

This means that 46 must equal 3x - 14

Note that we've just created an equation that we can use to solve for

We now use the equation to solve for x

3x - 14 = 46

add 14 to both sides

3x - 14 + 14 = 46 + 14

simplify

3x = 60

Divide both sides by 3

3x / 3 = x

60 / 3 = 20

We get that x = 20

4 0

3 years ago

PLZ HEEEELP!!!!! 10 POINTS!!!!

katrin2010 [14]

Answer:

Step-by-step explanation:

Part a

We need two inequalities, one for time worked at each job and the other

for amounts of money earned.

time: Let b and c represent the time (number of hours) worked at babysitting and landscaping respectively. Then b + c ≤ 20 hrs/wk

earnings: Let ($3/hr)(b) represents the amount of money earned babysitting for b hour. Let ($7/hr)(c) represent the money earned working at landscaping. These amounts are <em>per week</em>. The appropriate inequality is ($3/hr)(b) + ($7/hr)(c) ≥ $84 per week. The other inequality is

b + c ≤ 20 hrs/wk.

Part b:

As before, b + c ≤ 20 hrs/wk. What happens if Chet spends all his 20 hours babysitting? To answer this, set c = 0 (no landscaping hours). Then b ≤ 20 hours. At $3/hr, he could earn only $60 and have no time left for landscaping. Not good.

Let's experiment: suppose he works 15 hours babysitting and 5 hours landscaping. His earnings would be $45 + $35, or $80. Still not enough; he wants to earn $84 total. Let's redistribute his time and try again: suppose he works 14 hours babysitting and 6 hours landscaping; his earnings would be $42 + $42, or $84. So {b = 14 hours and c + 6 hours} is a solution. As we continue to reduce the number of hours Chet works babysitting and correspondingly increase those he works landscaping, his earnings will go up, beyond $84.

Here's a table that summarizes this:

babysitting landscaping total amount

hours hours earned

15 5 $60 (not acceptable)

14 6 $84 (borderline acceptable)

12 8 $36 + $42 = $78 (great)

6 14 $116 (greater still)

2 18 $132

1 19 $134

0 20 $140

Summary: Chet can work anywhere from 0 to 14 hours babysitting and expect to earn $84 or more.

4 0

3 years ago