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beks73 [17]
3 years ago
15

*nervous laugh* SO ummm geometry isnt my thing actually im rather bad it please help!

Mathematics
2 answers:
Savatey [412]3 years ago
8 0
I believe the answer would be D 92 and 88 hope it helped 
gregori [183]3 years ago
6 0
Te answer is C 100 and 80 straight lines equal out to be 180*
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A professional basketball team made 37.9% of its three-point field goals in one season. If 80 three-point field goal attempts ar
kompoz [17]

Answer:

The probability that more than 35 were made is 0.14686.

Step-by-step explanation:

We are given that a professional basketball team made 37.9% of its three-point field goals in one season.

80 three-point field goal attempts are randomly selected from the season.

Let \hat p = <em>sample proportion of three-point field goals made in one season.</em>

The z-score probability distribution for the sample proportion is given by;

                           Z  =  \frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n}} }  ~ N(0,1)

where, p = population proprotion of three-point field goals = 37.9% = 0.379

            n = sample of three-point field goal attempts = 80

            \hat p = sample proportion of three-point field goals = \frac{35}{80} = 0.4375

Now, if 80 three-point field goal attempts are randomly selected from the season, the probability that more than 35 were made is given by = P(\hat p > 0.4375)

P(\hat p > 0.4375) = P( \frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n}} } > \frac{0.4375-0.379}{\sqrt{\frac{0.4375(1-0.4375)}{80}} } ) = P(Z > 1.05) = 1 - P(Z \leq 1.05)

                                                                       = 1 - 0.85314 = <u>0.14686</u>

The above probability is calculated by looking at the value of x = 1.05 in the z table which has an area of 0.85314.

4 0
3 years ago
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