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3241004551 [841]
3 years ago
14

How long is a 1 foot pole?

Mathematics
2 answers:
Fudgin [204]3 years ago
6 0

it would be 12 inches. :)!!

STALIN [3.7K]3 years ago
4 0

Answer:

12 inches

    or

30.48 cm

Step-by-step explanation:

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Answer: A, because it is a 30, 60, 90 triangle. To get the 34 they had to times the number by 2 so you divide that by 2 and it gives you 17 then you have the square root of 3* 17, that doesn’t mathematically work so it stays as 17*the square root of 3
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3 years ago
Will mark brainliest
vovikov84 [41]
(0,0)-->(0,0)
(3,-1)-->(9,-3)
(3,3)-->(9,9)


so what you will plug after the dilation will be the points on the right/
hope i help
 
3 0
3 years ago
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What is new equation if the function y=2^x is translated 5 units to the right and 1 unit down?
Alja [10]
2^(x-5) -1 should be it
4 0
3 years ago
Birth weights of babies born to full-term pregnancies follow roughly a Normal distribution. At Meadowbrook Hospital, the mean we
Marina86 [1]

Answer:

Required Probability = 0.1283 .

Step-by-step explanation:

We are given that at Meadow brook Hospital, the mean weight of babies born to full-term pregnancies is 7 lbs with a standard deviation of 14 oz.

Firstly, standard deviation in lbs = 14 ÷ 16 = 0.875 lbs.

Also, Birth weights of babies born to full-term pregnancies follow roughly a Normal distribution.

Let X = mean weight of the babies, so X ~ N(\mu = 7 lbs , \sigma^{2}  = 0.875^{2}  lbs)

The standard normal z distribution is given by;

              Z = \frac{Xbar-\mu}{\frac{\sigma}{\sqrt{n} } } ~ N(0,1)

where, X bar = sample mean weight

             n = sample size = 4

Now, probability that the average weight of the four babies will be more than 7.5 lbs = P(X bar > 7.5 lbs)

P(X bar > 7.5) = P( \frac{Xbar-\mu}{\frac{\sigma}{\sqrt{n} } } > \frac{7.5-7}{\frac{0.875}{\sqrt{4} } }  ) = P(Z > 1.1428) = 0.1283 (using z% table)

Therefore, the probability that the average weight of the four babies will be more than 7.5 lbs is 0.1283 .

8 0
4 years ago
(a) Let {A1, A2} be a partition of a sample space and let B be any event. State and prove the Law of Total Probability as it app
Nataliya [291]

Answer:

0.625

Step-by-step explanation:

Given that {A1, A2} be a partition of a sample space and let B be any event. State and prove the Law of Total Probability as it applies to the partition {A1, A2} and the event B.

Since A1 and A2 are mutually exclusive and exhaustive, we can say

b) P(B) = P(A1B)+P(A2B)

Selecting any one coin is having probability 0.50. and A1, A2 are events that the coins show heads.P(B/A1) = 0.50 \\P(B/A2) = 0.75\\P(A1B) = 0.5(0.5) = 0.25 \\P(A2B) = 0.75(0.5) = 0.375\\P(B) = 0.625

c) Using Bayes theorem

conditional probability that it wasthe biased coin

=\frac{0.375}{0.625} \\=\frac{3}{5}

d) Given that the chosen coin flips tails,the conditional probability that it was the biased coin=\frac{0.25*0.5}{0.25*0.5+0.5*0.5} \\=\frac{1}{3}

7 0
3 years ago
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