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8090 [49]
3 years ago
13

the box plot suggests that about 25% of quarterbacks in the boys football league threw fewer than what number of touchdowns?

Mathematics
1 answer:
stiks02 [169]3 years ago
7 0

Answer: the answer is 7

Step-by-step explanation:

You might be interested in
Find the final amount of money in an account if $7, 200 is deposited at 2.5 % interest compounded
andreev551 [17]

The final amount is $7,615.27

A = P(1 + r/n)^t

Where,

A = Final amount

P = principal = $7, 200

r = interest rate = 2.5% = 0.025

n = number of periods = 4

t = time = 9 years

A = P(1 + r/n)^t

= 7,200(1 + 0.025/4)^9

= 7,200(1 + 0.00625)^9

= 7,200(1.00625)^9

= 7,200(1.0576769512798)

= 7,615.2740492152

Approximately,

A = $7,615.27

brainly.com/question/14003110

7 0
2 years ago
Use the given degree of confidence and sample data to construct a confidence interval for the population proportion p: n = 195,
Troyanec [42]

Answer:

c. 0.778 < p < 0.883.

Step-by-step explanation:

The formula for confidence interval for proportion =

p ± z score × √p(1 - p)/n

p = x/n

n = 195, x = 162

z score for 95% confidence Interval = 1.96

p = 162/195

p = 0.8307692308

p ≈ approximately equal to = 0.8308

0.8308 ± 1.96 × √0.8308 × (1 - 0.8308)/195

0.8308 ± 1.96 ×√0.8308 × 0.1692/195

0.8308 ± 1.96 × √0.0007208788

0.8308 ± 1.96 × 0.0268491862

0.8308 ± 0.052624405

Confidence Interval

= 0.8308 - 0.052624405

= 0.778175595

Approximately = 0.778

= 0.8308 + 0.052624405

= 0.883424405

Approximately = p

0.883

Therefore, the confidence interval for this proportion = (0.778, 0.883) or option c. 0.778 < p < 0.883

6 0
3 years ago
Now suppose that bigger cups are ordered and the machine’s mean amount dispensed is set at μ=12. Assuming we can precisely adjus
Elina [12.6K]

Answer:

σ should be adjusted at 0.5.

Step-by-step explanation:

The Empirical Rule states that, for a normally distributed random variable:

Approximately 68% of the measures are within 1 standard deviation of the mean.

Approximately 95% of the measures are within 2 standard deviations of the mean.

Approximately 99.7% of the measures are within 3 standard deviations of the mean.

In this problem, we have that:

Mean 12.

Assuming we can precisely adjust σ, what should we set σtobe so that the actual amount dispensed is between 11 and 13 ounces, 95% of the time?

13 should be 2 standard deviations above the mean of 12, and 11 should be two standard deviations below the mean.

So 1 should be worth two standard deviations. Then

2\sigma = 1

\sigma = \frac{1}{2}

\sigma = 0.5

σ should be adjusted at 0.5.

4 0
2 years ago
May somebody please help me and actually not just type nonsense I’ll give Brainlest or whatever u want
Lerok [7]
I’m pretty sure that the figure is a rectangle and the two longer sides are 9 centimeters and the shorter ones are 6 centimeters. Rectangles have 4 right angles and all of the side lengths added together equals 30 centimeters
4 0
3 years ago
Read 2 more answers
I need some help I have no idea <br><br> 22L/min= xL/hour <br> Complete the conversion.
omeli [17]

Answer:

may be 1320L/hr

Step-by-step explanation:

not sure

7 0
3 years ago
Read 2 more answers
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