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laila [671]
2 years ago
7

4.8/2z+15=0.2/5 Plz help I am only 8

Mathematics
2 answers:
kakasveta [241]2 years ago
8 0

Answer:  z = 247.5

Step-by-step explanation:

4.8/2z+15=0.2/5 To eliminate the denominators, multiply both sides by their reciprocals.

2z(5)(4.8/2z+15)=2z(5)(0.2/5)  Denominators "cancel." Distribute 5(4.8 + 15)

24 + 75 = 2z(0.2)    Simplify: Add  24+75=99 and Multiply 2z×0.2= 0.4z

99 = 0.4z .  Divide both sides by 0.4

99/0.4 = 0.4z/0.4 .   0.4  "cancels" on the right. dividing by a decimal is like multiplying by its reciprocal, 10/4 × 99

247.5 = z

yaroslaw [1]2 years ago
3 0

Answer:

z=−6.233333

Step-by-step explanation:

4.8

2

z+15=

0.2

5

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Oatmeal is packaged in a cylindrical container, as shown in the diagram. The diameter of the container is 13 centimeters and its
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5 0
3 years ago
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).

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(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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