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baherus [9]
3 years ago
8

Write an EQUATION that represents what is seen in the figure.

Mathematics
1 answer:
Ostrovityanka [42]3 years ago
7 0

Step-by-step explanation:

x°+55°+40°=180°

x=180-95

<h3> ×=85°</h3>
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Rita read a book with 973 pages in one week. if she read the same number of pages each day how many pages did rita read each day
Ede4ka [16]
973 pages divided by 7 days in the week equals 139 pages each day.

5 0
3 years ago
One measure of an athlete’s ability is the height of his or her vertical leap. Many professional basketball players are known fo
almond37 [142]

Answer:

(1) P(\bar X < 26 inches) = 0.0436

(2) P(27.5 inches < \bar X < 28.5 inches) = 0.2812

Step-by-step explanation:

We are given that the mean vertical leap of all NBA players is 28 inches. Suppose the standard deviation is 7 inches and 36 NBA players are selected at random.

Firstly, Let \bar X = mean vertical leap for the 36 players

Assuming the data follows normal distribution; so the z score probability distribution for sample mean is given by;

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

where, \mu = population mean vertical  leap = 28 inches

            \sigma = standard deviation = 7 inches

            n = sample of NBA player = 36

(1) Probability that the mean vertical leap for the 36 players will be less than 26 inches is given by = P(\bar X < 26 inches)

   P(\bar X < 26) = P( \frac{\bar X - \mu}{\frac{\sigma}{\sqrt{n} } } < \frac{26-28}{\frac{7}{\sqrt{36} } } ) = P(Z < -1.71) = 1 - P(Z \leq 1.71)

                                                 = 1 - 0.95637 = 0.0436

(2) <em>Now, here sample of NBA players is 26 so n = 26.</em>

Probability that the mean vertical leap for the 26 players will be between 27.5 and 28.5 inches is given by = P(27.5 inches < \bar X < 28.5 inches) = P(\bar X < 28.5 inches) - P(\bar X \leq 27.5 inches)

    P(\bar X < 28.5) = P( \frac{\bar X - \mu}{\frac{\sigma}{\sqrt{n} } } < \frac{28.5-28}{\frac{7}{\sqrt{26} } } ) = P(Z < 0.36) = 0.64058 {using z table}                      

    P(\bar X \leq 27.5) = P( \frac{\bar X - \mu}{\frac{\sigma}{\sqrt{n} } } \leq \frac{27.5-28}{\frac{7}{\sqrt{26} } } ) = P(Z \leq -0.36) = 1 - P(Z < 0.36)

                                                        = 1 - 0.64058 = 0.35942

Therefore, P(27.5 inches < \bar X < 28.5 inches) = 0.64058 - 0.35942 = 0.2812

6 0
4 years ago
A 5 inch x 7 inch photograph is placed inside a picture frame. Both the length and width of the frame are 2x inches larger than
Natasha2012 [34]
It should be 10 inch x 14 inch if it is 2x larger
5 0
3 years ago
Write a statement that indicates that the triangles in each pair are congruent.
Murljashka [212]

Answer:

Step-by-step explanation:

6 0
3 years ago
According to exit polling from the 2014 U.S. midterm elections, 36% of voters had a household income less than $50,000, while 64
Veseljchak [2.6K]

Answer:

0.6946 = 69.46% probability that a randomly selected Republican voter from the exit poll is from a household that makes at least $50,000.

Step-by-step explanation:

Conditional Probability

We use the conditional probability formula to solve this question. It is

P(B|A) = \frac{P(A \cap B)}{P(A)}

In which

P(B|A) is the probability of event B happening, given that A happened.

P(A \cap B) is the probability of both A and B happening.

P(A) is the probability of A happening.

In this question:

Event A: Republican

Event B: From a household that makes at least $50,000.

Probability of Republican:

43% of 36%(makes less than $50,000).

55% of 64%(makes more than $50,000).

So

P(A) = 0.43*0.36 + 0.55*0.64 = 0.5068

Republican and from a household that makes at least $50,000.

55% of 64%. So

P(A \cap B) = 0.55*0.64 = 0.352

What is the probability that a randomly selected Republican voter from the exit poll is from a household that makes at least $50,000?

P(B|A) = \frac{P(A \cap B)}{P(A)} = \frac{0.352}{0.5068} = 0.6946

0.6946 = 69.46% probability that a randomly selected Republican voter from the exit poll is from a household that makes at least $50,000.

4 0
3 years ago
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