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2 years ago

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Suzanne owns a small business that employs 555 other people. Suzanne makes \$100{,}000$100,000dollar sign, 100, comma, 000 per y

ear, and the other 555 employees make between \$40{,}000$40,000dollar sign, 40, comma, 000 and \$50{,}000$50,000dollar sign, 50, comma, 000 per year.
Suzanne decides to increase her salary by \$30{,}000$30,000dollar sign, 30, comma, 000 per year and leave the rest of the salaries the same.
How will increasing her salary affect the mean and median? How will increasing her salary affect the mean and median?

2 answers:

zalisa [80]2 years ago

6 0

Answer:

16

Step-by-step explanation:

Tpy6a [65]2 years ago

6 0

Answer:

Mean increased

Median remains same

Step-by-step explanation:

Suzanne make $ 100 k per year

5 other employee $ 40 k to $ 50 K per year per employee

data in ascending order

$40000 , $40000+a , $40000+b , $40000+c , $50000 , $ 100000

where 10000 ≥c ≥ b ≥ a ≥ 0

Mean = ($40000 + $40000+a + $40000+b + $40000+c + $50000 + $ 100000) /6

Mean = $(310000 + a + b + c)/ 6

Median = ($40000+b + $40000+c )/2 = $40000 + (b+c)/2

Suzanne make now $ 100000 + $ 30000 = $ 130000 per year

$40000 , $40000+a , $40000+b , $40000+c , $50000 , $ 130000

Mean = ($40000 + $40000+a + $40000+b + $40000+c + $50000 + $ 130000) /6

Mean = $(340000 + a + b + c)/ 6

Median = ($40+b k + $40+c k)/2 = $40 + (b+c)/2 k

Mean increased by $(340000 + a + b + c)/ 6 - $(310000 + a + b + c)/ 6

= $ 30000/6 = $ 5000

Median Remains the same

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A farmer's annual income is represented by the variable x. The farmer will pay 25% of his income for federal taxes and 9% of his

nevsk [136]

Answer:

D) x - 0.34x

Step-by-step explanation:

25% + 9% = 34% = 0.34

Will pay 34% in taxes, means he subtracts from is annual income.

x - 0.34x

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2 years ago

Test the given claim. Identify the null hypothesis, alternative hypothesis, test statistic, P-value, and then state the concl

never [62]

Answer:

a) Failed to reject the null hypothesis (P-value=0.09).

b) The 95% CI for the difference in proportions is:

$-0.0599\leq\pi_1-\pi_2\leq0.0124$

Step-by-step explanation:

a) We have to perform a hypothesis test for the difference of proportions.

The null and alternative hypothesis are:

$H_0: \pi_1\geq\pi_2\\\\H_1: \pi_1$

The significance level is 0.05.

The proportion of the passenger cars owners is:

$p_1=\frac{239}{2142} =0.1116$

The proportion of commercial truck owners is:

$p_2=\frac{54}{399}=0.1353$

The weigthed average p is

$p=\frac{n_1p_1+n_2p_2}{n_1+n_2}=\frac{239+54}{2142+399}=0.1153$

The estimated standard deviation is

$s=\sqrt{\frac{p(1-p)}{n_1}+\frac{p(1-p)}{n_2}} =\sqrt{\frac{0.1153(1-0.1153)}{2142}+\frac{0.1153(1-0.1153)}{399}} =0.0174$

We can calculate the z-value as:

$z=\frac{\Delta p}{s}=\frac{0.1116-0.1353}{0.0174}=-1.362$

The P-value for z=-1.362 is P=0.0866.

The P-value (0.09) is greater than the significance level (0.05), so it failed to reject the null hypothesis. There is no enough evidence to prove that commercial trucks owners violate laws requiring front license plates at a higher rate than owners of passenger cars.

b) We can construct a 95% CI, according to the significance level of 0.05.

The z-value for this CI is 1.96.

We have to recalculate the standard deviation:

$\sigma=\sqrt{\frac{p_1(1-p_1)}{n_1} +\frac{p_2(1-p_2)}{n_2}} =\sqrt{\frac{0.1116(1-0.1116)}{2142} +\frac{0.1353(1-0.1353)}{399}} =0.0184$

The lower limit is then:

$LL=(p_1-p_2)-z*\sigma=(0.1116-0.1353)-1.96*0.0184=-0.0238-0.0361\\\\LL=-0.0599$

The upper limit is:

$UL=(p_1-p_2)+z*\sigma=(0.1116-0.1353)+1.96*0.0184=-0.0238+0.0361\\\\UL=0.0124$

The 95% CI for the difference in proportions is:

$-0.0599\leq\pi_1-\pi_2\leq0.0124$

In this case, we can conclude that the difference between the proportions, with 95% confidence, can still be equal or greater than zero, meaning that it is possible passenger car owners violate laws more than truck owners.

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2 years ago

Suppose the time a child spends waiting at for the bus as a school bus stop is exponentially distributed with mean 7 minutes. De

Gala2k [10]

Answer:

The probability that the child must wait between 6 and 9 minutes on the bus stop on a given morning is 0.148.

Step-by-step explanation:

Let the random variable <em>X</em> represent the time a child spends waiting at for the bus as a school bus stop.

The random variable <em>X</em> is exponentially distributed with mean 7 minutes.

Then the parameter of the distribution is, $\lambda=\frac{1}{\mu}=\frac{1}{7}$ .

The probability density function of <em>X</em> is:

$f_{X}(x)=\lambda\cdot e^{-\lambda x};\ x>0,\ \lambda>0$

Compute the probability that the child must wait between 6 and 9 minutes on the bus stop on a given morning as follows:

$P(6\leq X\leq 9)=\int\limits^{9}_{6} {\lambda\cdot e^{-\lambda x}} \, dx$

$=\int\limits^{9}_{6} {\frac{1}{7}\cdot e^{-\frac{1}{7} \cdot x}} \, dx \\\\=\frac{1}{7}\cdot \int\limits^{9}_{6} {e^{-\frac{1}{7} \cdot x}} \, dx \\\\=[-e^{-\frac{1}{7} \cdot x}]^{9}_{6}\\\\=e^{-\frac{1}{7} \cdot 6}-e^{-\frac{1}{7} \cdot 9}\\\\=0.424373-0.276453\\\\=0.14792\\\\\approx 0.148$

Thus, the probability that the child must wait between 6 and 9 minutes on the bus stop on a given morning is 0.148.

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2 years ago

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vfiekz [6]

Yes. The original price of the dozen added to the price of berries (.5) multiplied by the amount(b) is the same equational representation to the right of the equal sign.

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2 years ago

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Landon babysits and works part time at tge water park over the summer. onw week, he babysat for 3 hours and worked at the water

Tems11 [23]

B = hourly rate for babysitting and w = hourly rate for working at water park

3b + 10w = 109...multiply by -8
8b + 12w = 177...multiply by 3
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-24b - 80w = - 872 (result of multiplying by -8)
24b + 36w = 531 (result of multiplying by 3)
---------------------add
- 44w = - 341
w = -341/-44
w = 7.75 <=== hourly rate for working at water park

3b + 10w = 109
3b + 10(7.75) = 109
3b + 77.50 = 109
3b = 109 - 77.50
3b = 31.50
b = 31.50/3
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2 years ago

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