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sdas [7]
3 years ago
13

A football player completes a pass 69.4​% of the time. Find the probability that​ (a) the first pass he completes is the second​

pass, (b) the first pass he completes is the first or second​ pass, and​ (c) he does not complete his first two passes.
Mathematics
1 answer:
wlad13 [49]3 years ago
6 0

Step-by-step explanation:

(a) If his second pass is the first that he completes, that means he doesn't complete his first pass.

P = P(not first) × P(second)

P = (1 − 0.694) (0.694)

P ≈ 0.212

(b) This time we're looking for the probability that he doesn't complete the first but does complete the second, or completes the first and not the second.

P = P(not first) × P(second) + P(first) × P(not second)

P = (1 − 0.694) (0.694) + (0.694) (1 − 0.694)

P ≈ 0.425

(c) Finally, we want the probability he doesn't complete either pass.

P = P(not first) × P(not second)

P = (1 − 0.694) (1 − 0.694)

P ≈ 0.094

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Does anyone know the answer to this question.....0.5x + 1.75 = 10
marysya [2.9K]

Answer:

x = 16.5

Step-by-step explanation:

0.5x + 1.75 = 10

Subtract 1.75 from both sides:

0.5x = 8.25

Divide both sides by 0.5

x = 16.5

Check your work!

0.5(16.5) + 1.75 = 10

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3 years ago
A lumber company is making doors that are 2058.0 millimeters tall. If the doors are too long they must be trimmed, and if the do
USPshnik [31]

Answer:

No, we don't have sufficient evidence to support the claim that the doors are either too long or too short.

Step-by-step explanation:

We are given that a lumber company is making doors that are 2058.0 millimeters tall. If the doors are too long they must be trimmed, and if the doors are too short they cannot be used.

A sample of 11 is made, and it is found that they have a mean of 2043.0 millimeters with a standard deviation of 25.0.

Let \mu = <u><em>population mean length of doors</em></u>.

So, Null Hypothesis, H_0 : \mu = 2058.0 millimeters      {means that the mean length of doors is 2058.0 millimeters}

Alternate Hypothesis, H_A : \mu\neq 2058.0 millimeters     {means that the doors are either too long or too short}

The test statistics that will be used here is <u>One-sample t-test statistics</u> because we know about the population standard deviation;

                               T.S.  =  \frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }  ~ t_n_-_1

where, \bar X = sample mean = 2043.0 millimeters

             s = sample standard deviation = 25.0 millimeters

             n = sample of doors = 11

So, <u>the test statistics</u> =  \frac{2043.0-2058.0}{\frac{25.0}{\sqrt{11} } }

                                     =  -1.989  

The value of t-test statistics is -1.989.

Now, at a 5% level of significance, the t table gives a critical value between -2.228 and 2.228 at 10 degrees of freedom for the two-tailed test.

Since the value of our test statistics lies within the range of critical values of t, so <u><em>we have insufficient evidence to reject our null hypothesis</em></u> as it will not fall in the rejection region.

Therefore, we conclude that the mean length of doors is 2058.0 millimeters.

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3 years ago
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5 2/5 is the answer as a mixed number
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Iteru [2.4K]

Answer:

500

Step-by-step explanation:

15% * x = 75

(15/100) * x = 75           Multiply both sides by 100/15

(15/100)*(100/15) = 75 * 100/15

x = 7500/15

x = 500

Answer 500

Proportion

15/100 = 75 / x             Cross multiply

15x = 100*75                Combine the right

15x = 7500                   Divide 15

15x/15 = 7500/15

x = 500


3 0
3 years ago
Read 2 more answers
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