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3 years ago

Math Problem again will mark brainliest.

Mathematics

2 answers:

Brums [2.3K]3 years ago

5 0

Answer:

The answer to this problem would be A.

Step-by-step explanation:

For childcare facilities for each visit is $9, we would say 9x because we dont know the exact amount of time that she visits the gym with her child. Her monthly gym bill is $32 which would ve added to the amount she would pay for childcare. The sum of it all should be less than or equal to that of $122.

katen-ka-za [31]3 years ago

4 0

Answer:

A. 9x + 32 ≤ 122

Step-by-step explanation:

$32 is the fixed monthly due

$9 everytime she uses childcare

$122 is the maximum she wants to spend This month.

32 + 9x ≤ 122

You might be interested in

-6p + 43=13 What is p?

son4ous [18]

Subtract 43 from both sides
-6p=-30

Divide both sides by -6
p=5

Final answer: p=5

7 0

3 years ago

Maybe brainliest if its right

ahrayia [7]

Answer:

2.5 cm

Step-by-step explanation:

5 Cm is the diameter and the radius is half the diameter so 5/2=2.5 or 1/2

4 0

3 years ago

Read 2 more answers

What is the value of the expression below when X=48 and Y=-5 A:2 B:8.65 C:22 D:-2

GrogVix [38]

A is the answer

Hope this helps

3 0

2 years ago

Can someone answer this question please answer it correctly if it’s corect I will mark you brainliest

gogolik [260]

Answer:

7. 21 times more

8 1/9

Step-by-step explanation:

7. 2 pants, 5 tops , 3 gloves, 7 masks

2*5*3*7 =210 outfits

10 * what = 210

10 *21 = 210

21 times more

8. There are 36 possible outcomes

11 121 131 14 15 16

21 22 23 24 25 26

31 31 33 34 35 36

41 42 43 44 45 46

51 52 53 54 55 56

61 62 63 64 65 66

There are 36 possibilities and 4 have a sum of 9

P (sum of 9) = outcomes with sum of 9/total

4/36 = 1/9

6 0

3 years ago

A sales and marketing management magazine conducted a survey on salespeople cheating on their

Alenkinab [10]

Answer:

The 95% confidence interval estimate of the population proportion of managers who have caught salespeople cheating on an expense report is (0.5116, 0.6484).

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of $\pi$ , and a confidence level of $1-\alpha$ , we have the following confidence interval of proportions.

$\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}$

In which

z is the zscore that has a pvalue of $1 - \frac{\alpha}{2}$ .

In the survey on 200 managers, 58% of the managers have caught salespeople cheating on an expense report.

This means that $n = 200, \pi = 0.58$

95% confidence level

So $\alpha = 0.05$ , z is the value of Z that has a pvalue of $1 - \frac{0.05}{2} = 0.975$ , so $Z = 1.96$ .

The lower limit of this interval is:

$\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.58 - 1.96\sqrt{\frac{0.58*0.42}{200}} = 0.5116$

The upper limit of this interval is:

$\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.58 + 1.96\sqrt{\frac{0.58*0.42}{200}} = 0.6484$

The 95% confidence interval estimate of the population proportion of managers who have caught salespeople cheating on an expense report is (0.5116, 0.6484).

7 0

3 years ago