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lawyer [7]
3 years ago
6

HELP!!! 15 points and marked brainliest

Mathematics
1 answer:
AVprozaik [17]3 years ago
7 0

Answer:

35.8 I just used a calculator to find the answer

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A high school basketball coach is selecting his team. The minimum and maximum height requirements are as follows:
daser333 [38]

Answer:

B. About 2% of the boys are eligible to be a small forward on the team

Step-by-step explanation:

Recall : 1 feets = 12 inches

Point guard = 6’2" – 6’6" tall = 74 - 78 inches

Mean = 70 ; Standard deviation = 4

Z = (x - mean) / standard

P(x < 74) = (74 - 70) / 4 = 1

P(x < 78) = (78 - 70) / 4 = 2

0.97725 - 0.84134 = 0.13591

Small forward : 6'6" = 78 inches

P(x ≥ 78) = (78 - 70) / 4 = 2

P(z ≥ 2) = 0.02275 = 2.275% about 2%

Centre : 6'8" = 80

P(x ≥ 80) = (80 - 70) / 4 = 2.5

P(z ≥ 2.5) = 0.0062097 = 0.62%

8 0
3 years ago
What is the area of the parallelogram QRST?
jonny [76]

\text{Area of parallelogram QRST} =  QR \times TU = 20 ~in \times 18~in =360 in^2

8 0
2 years ago
PLZ I WILL GIVE BRAINLIEST AND IT HAS 20 POINTS The volume of a cylinder is approximately 72 feet cubed. Which is the best appro
solong [7]

Answer:

24 feet cubed

Step-by-step explanation:

When comparing the volume formula for a cone and the volume formula for a cylinder, a cone is 1/3 of a cylinder. Therefore, divide 72 by 3 to get the volume of a cone with the same base and height. It would not be 24 pi feet cubed because 72 feet cubed is already multiplied by the pi.

3 0
3 years ago
Read 2 more answers
Does the relationship in this table represent a proportional relationship? x 0.25 0.5 0.6 1.75 y 3.25 3.5 3.6 4.75 Select from t
ioda

Answer:

yes it does I had this question

4 0
3 years ago
Read 2 more answers
A population of values has a normal distribution with μ=204.9μ=204.9 and σ=81.9σ=81.9. You intend to draw a random sample of siz
marin [14]

Answer:

7. μ=204.9 and σ=5.4968

8. μ=75.9 and σ=0.7136

9. p=0.9452

Step-by-step explanation:

7. - Given that the population mean =204.9 and the standard deviation is 81.90 and the sample size n=222.

-The sample mean,\mu_xis calculated as:

\mu_x=\mu=204.9, \mu_x=sample \ mean

-The standard deviation,\sigma_x is calculated as:

\sigma_x=\frac{\sigma}{\sqrt{n}}\\\\=\frac{81.9}{\sqrt{222}}\\\\=5.4968

8. For a random variable X.

-Given a X's population mean is 75.9, standard deviation is 9.6 and a sample size of 181

-#The sample mean,\mu_x is calculated as:

\mu_x=\mu\\\\=75.9

#The sample standard deviation is calculated as follows:

\sigma_x=\frac{\sigma}{\sqrt{n}}\\\\=\frac{9.6}{\sqrt{181}}\\\\=0.7136

9. Given the population mean, μ=135.7 and σ=88 and n=59

#We calculate the sample mean;

\mu_x=\mu=135.7

#Sample standard deviation:

\sigma_x=\frac{\sigma}{\sqrt{n}}\\\\=\frac{88}{\sqrt{59}}\\\\=11.4566

#The sample size, n=59 is at least 30, so we apply Central Limit Theorem:

P(\bar X>117.4)=P(Z>\frac{117.4-\mu_{\bar x}}{\sigma_x})\\\\=P(Z>\frac{117.4-135.7}{11.4566})\\\\=P(Z>-1.5973)\\\\=1-0.05480 \\\\=0.9452

Hence, the probability of a random sample's mean being greater than 117.4 is 0.9452

7 0
3 years ago
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