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Vilka [71]
3 years ago
7

Will mark brainliest just plz answer

Mathematics
1 answer:
sveticcg [70]3 years ago
5 0

Answer:

(0,2)

Step-by-step explanation:

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Solve the formula for the variable h f=12gh
Contact [7]
To solve for variable h flip the equation. You get: 12gh=f. After, divide both sides by 12 to get your answer: h=f/12g
8 0
3 years ago
I will give brainliest
Irina18 [472]

Answer: -3

Step-by-step explanation:

8 0
2 years ago
Solve for x in simplest form.<br>14 = 1/3(8x + 6)​
elena-s [515]

Answer:

x = 4.5

Step-by-step explanation:

14 = 8/3x + 2

12 = 8/3x

36 = 8x

x = 4.5

5 0
3 years ago
Suppose the probability of an IRS audit is 1.5 percent for U.S. taxpayers who file form 1040 and who earned $100,000 or more.
sp2606 [1]

Answer:

(A) The odds that the taxpayer will be audited is approximately 0.015.

(B) The odds against these taxpayer being audited is approximately 65.67.

Step-by-step explanation:

The complete question is:

Suppose the probability of an IRS audit is 1.5 percent for U.S. taxpayers who file form 1040 and who earned $100,000 or more.

A. What are the odds that the taxpayer will be audited?

B. What are the odds against such tax payer being audited?

Solution:

The proportion of U.S. taxpayers who were audited is:

P (A) = 0.015

Then the proportion of U.S. taxpayers who were not audited will be:

P (A') = 1 - P (A)

        = 1 - 0.015

        = 0.985

(A)

Compute the  odds that the taxpayer will be audited as follows:

\text{Odds of being Audited}=\frac{P(A)}{P(A')}

                                    =\frac{0.015}{0.985}\\\\=\frac{3}{197}\\\\=0.015228\\\\\approx 0.015

Thus, the odds that the taxpayer will be audited is approximately 0.015.

(B)

Compute the odds against these taxpayer being audited as follows:

\text{Odds against Audited}=\frac{P(A')}{P(A)}

                                    =\frac{0.985}{0.015}\\\\=\frac{3}{197}\\\\=65.666667\\\\\approx 65.67

Thus, the odds against these taxpayer being audited is approximately 65.67.

8 0
4 years ago
The average fuel efficiency of U.S. light vehicles for 2005 was 21 mpg. If the standard deviation of the population was 2.9 and
statuscvo [17]

Answer:

1) The probability that the mean mpg for a random sample of 25 light vehicles is 0.042341.

2) between 20 and 25 --> 21-25/2.9 = -1.38

Step-by-step explanation:

Problem #1:

  • Using the z-score formula, z = (x-μ)/σ/n, where x is the raw score = 20 mpg,μ is the population mean = 21 mpg , σ is the population standard deviation = 2.9, n = random number of samples.
<h3><u>X < 20</u></h3>
  • = z = 20 - 21/2.9/√25
  • = z = -1/2.9/5
  • = z = -1.72414
<h2><u><em>Now</em></u></h2>

<em>P-value from Z-Table:</em>

<h3><u>P(x<20) = 0.042341</u></h3>

Problem #2:

<h3>21-25/2.9 = -1.38</h3>
6 0
3 years ago
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