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elena-s [515]
3 years ago
9

Define the term net income .. according to the given context​

Mathematics
1 answer:
Pachacha [2.7K]3 years ago
4 0

ANSWER: the amount an individual earns after subtracting all expenses, taxes and costs.

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Please help!! Thanks
Anna35 [415]
The answer is 85.

Substitute 3 for x and 4 for y in the expression.

After simplifying using order of operations, you will get 85:)
4 0
3 years ago
If f(x)= 3/x+2 - square root of x-3, the domain for f(x) is all real numbers ___ than or equal to 3
aksik [14]

Answer:

I don't get the question very well

5 0
3 years ago
Read 2 more answers
Hadlee got 18 out of 20 questions correct on her math quiz. What percent of the questions did she get right?
adelina 88 [10]
It would be a 90%. In the calculator 18 is what percent of 20.
4 0
3 years ago
Read 2 more answers
PLEASE HELP I DON'T UNDERSTAND! A florist is making regular bouquets and mini bouquets. The florist has 118 roses and 226 peonie
Anna [14]

Answer:

The florist can make 11 regular bouquets and 21 mini bouquets

Step-by-step explanation:

The number of roses the florist has = 118 roses

The number of peonies the florist has = 226 peonies

The number of roses in each regular bouquet = 5 roses

The number of peonies in each regular bouquet = 11 peonies

The number of roses in each mini bouquet = 3 roses

The number of peonies in each mini bouquet = 5 peonies

1. The equation that gives the total number of roses to be used in both kinds of bouquet is given as follows;

Let r represent the number of regular bouquet the florist can make and let m represent the number of mini bouquet the florist can make, we have;

5·r + 3·m = 118...(1)

2. Similarly, we have;

11·r + 5·m  = 226...(2)

Making m the subject of the formula of both equations, and equating both values of m to find a common solution, we have;

m = (118 - 5·r)/3

m = (226 - 11·r)/5

(118 - 5·r)/3 = (226 - 11·r)/5

5 × (118 - 5·r) = 3 × (226 - 11·r)

590 - 25·r = 678 - 33·r

33·r - 25·r = 678 - 590 = 88

8·r = 88

r = 88/8 = 11

r = 11

The number of regular bouquet the florist can make = r = 11

m = (118 - 5·r)/3 = (118 - 5×11)/3 = 21

m = 21

The number of mini bouquet the florist can make = m = 21

The number of regular bouquet the florist can make = 11 bouquets

The number of mini bouquet the florist can make = 21 bouquets.

3 0
2 years ago
Experian would like to test the hypothesis that the average credit score for an adult in Virginia is different from the average
aliya0001 [1]

Answer:

a. We fail to reject the hypothesis that the average credit score for an adult in Virginia is different from the average credit score for an adult in North Carolina at the significance level of 0.05.

b. The 95% confidence interval for the true difference of means is -2.2468 and 36.2468. There is a probability of 95% that the true difference of means \mu_{1}-\mu_{2} is between -2.2468 and 36.2468. This confidence interval contains the number 0, this is consistent with the results we got in a. Because we fail to reject the hypothesis that the average credit score for an adult in Virginia is different from the average credit score for an adult in North Carolina at the significance level of 0.05.

Step-by-step explanation:

Let \mu_{1}-\mu_{2} be the true difference between the average credit score for an adult in Virginia and the average credit score for an adult in North Carolina. We have the large sample sizes n_{1} = 40 and n_{2} = 35, the unbiased point estimate for \mu_{1}-\mu_{2} is \bar{x}_{1} - \bar{x}_{2}, i.e., 699-682 = 17.

The standard error is given by \sqrt{\frac{\sigma^{2}_{1}}{n_{1}}+\frac{\sigma^{2}_{2}}{n_{2}}}, i.e.,

\sqrt{\frac{(44)^{2}}{40}+\frac{(41)^{2}}{35}} = 9.8198.

a. We want to test H_{0}: \mu_{1}-\mu_{2} = 0 vs H_{1}: \mu_{1}-\mu_{2} \neq 0 (two-tailed alternative). The rejection region is given by RR = {z | z < -1.96 or z > 1.96} where -1.96 and 1.96 are the 2.5th and 97.5th quantiles of the standard normal distribution respectively. The test statistic is Z = \frac{\bar{x}_{1} - \bar{x}_{2}-0}{\sqrt{\frac{\sigma^{2}_{1}}{n_{1}}+\frac{\sigma^{2}_{2}}{n_{2}}}} and the observed value is z_{0} = \frac{17}{9.8198} = 1.7312. Because 1.7312 does not fall inside RR, we fail to reject the null hypothesis.

b. The endpoints for a 95% confidence interval for \mu_{1}-\mu_{2} is given by 17\pm (z_{0.05/2})9.8198, i.e., 17\pm (z_{0.025})9.8198 where z_{0.025} is the 2.5th quantile of the standard normal distribution, i.e., -1.96, so, we have 17-(1.96)(9.8198) and 17+(1.96)(9.8198), i.e., -2.2468 and 36.2468. There is a probability of 95% that the true difference of means \mu_{1}-\mu_{2} is between -2.2468 and 36.2468. This confidence interval contains the number 0, this is consistent with the results we got in a. Because we fail to reject the hypothesis that the average credit score for an adult in Virginia is different from the average credit score for an adult in North Carolina at the significance level of 0.05.

3 0
3 years ago
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