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Rom4ik [11]
2 years ago
6

Geometry help needed. more information in the picture. please help. would be greatly appreciated.

Mathematics
1 answer:
ratelena [41]2 years ago
4 0

Answer:

60

Step-by-step explanation:

180-120=60so the answer is a 60

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Quantity of gasoline needed by a car to run 800 miles = 30 gallons

Quantity of gasoline needed by a car to run 1 mile =

= 30 ÷ 800

= 0.0375 gallons

So , to run 1 mile a car would need = 0.0375 gallons of oil

To run 700 miles the quantity of gasoline needed =

= 700 × 0.0375

= 26.25 gallons of gasoline

Therefore , a car will use 26.25 gallons of gasoline on a trip of 700 miles .

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Solve each of the following quadratic equations.
defon
A. x = 5
b. x = 3.38…
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For which value of x does each expression make sense
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A simple random sample of size nequals10 is obtained from a population with muequals68 and sigmaequals15. ​(a) What must be true
valentina_108 [34]

Answer:

(a) The distribution of the sample mean (\bar x) is <em>N</em> (68, 4.74²).

(b) The value of P(\bar X is 0.7642.

(c) The value of P(\bar X\geq 69.1) is 0.3670.

Step-by-step explanation:

A random sample of size <em>n</em> = 10 is selected from a population.

Let the population be made up of the random variable <em>X</em>.

The mean and standard deviation of <em>X</em> are:

\mu=68\\\sigma=15

(a)

According to the Central Limit Theorem if we have a population with mean <em>μ</em> and standard deviation <em>σ</em> and we take appropriately huge random samples (<em>n</em> ≥ 30) from the population with replacement, then the distribution of the sample mean will be approximately normally distributed.

Since the sample selected is not large, i.e. <em>n</em> = 10 < 30, for the distribution of the sample mean will be approximately normally distributed, the population from which the sample is selected must be normally distributed.

Then, the mean of the distribution of the sample mean is given by,

\mu_{\bar x}=\mu=68

And the standard deviation of the distribution of the sample mean is given by,

\sigma_{\bar x}=\frac{\sigma}{\sqrt{n}}=\frac{15}{\sqrt{10}}=4.74

Thus, the distribution of the sample mean (\bar x) is <em>N</em> (68, 4.74²).

(b)

Compute the value of P(\bar X as follows:

P(\bar X

                    =P(Z

*Use a <em>z</em>-table for the probability.

Thus, the value of P(\bar X is 0.7642.

(c)

Compute the value of P(\bar X\geq 69.1) as follows:

Apply continuity correction as follows:

P(\bar X\geq 69.1)=P(\bar X> 69.1+0.5)

                    =P(\bar X>69.6)

                    =P(\frac{\bar X-\mu_{\bar x}}{\sigma_{\bar x}}>\frac{69.6-68}{4.74})

                    =P(Z>0.34)\\=1-P(Z

Thus, the value of P(\bar X\geq 69.1) is 0.3670.

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F(x)=3/5x+1 find f(-15)
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