Based on the distance of the pitcher's mound from the home plate, the path of the ball, and the height the ball was hit, the distance the outfielder threw the ball is C. 183.0 ft.
<h3>How far did the outfielder throw the ball?</h3>
Based on the shape of a mound, the law of cosines can be used.
The distance the ball was thrown by the outfielder can therefore be d.
Distance is:
d² = 60.5² + 226² - (2 x 60.5 x 226 x Cos(39))
d ²= 33,484.42ft
Then find the square root:
d = √33,484.42
= 182.9874
= 183 ft
Find out more solving for distance at brainly.com/question/10739348
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Answer:
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Step-by-step explanation:
Step-by-step explanation:
It would be c plz mark me brainliest
Answer:
12
Step-by-step explanation:
To have an even integer the number's endian has to be an even number. Consider the 2-digit positive number like _ _ and place the number of possible options out of four total options in each blank space. We have 3 even options, so we can fill the endian blank like this _ <u>3</u>
Since the first digit can be any number, therefor, we have 4 options for the first blank space <u>4</u> <u>3</u>. Just multiply the two numbers and you have the answer.
22, 24, 28, 42, 44, 48, 52, 54, 58, 82, 84, 88
Answer:
Step-by-step explanation:
Hello!
X: the lifespan of a new computer monitor of Glotech.
The average life is μ= 85 months and the variance δ²= 64
And a sample of 122 monitors was taken.
You need to calculate the probability that the sample mean is greater than 86.6 months.
Assuming that the variable has a normal distribution X~N(μ;δ²), then the distribution of the sample mean is X[bar]~N(μ;δ²/n)
To calculate this probability you have to work using the sampling distribution and the following formula Z= (X[bar]-μ)/δ/√n ~N(0;1)
P(X[bar]>86.6)= 1 - P(X[bar]≤86.6)
1 - P(Z≤(86.6-85)/(8/√122))= 1 - P(Z≤2.21)= 1 - 0.98645= 0.013355
The probability of the sample mean is greater than 0.013355
I hope this helps!
