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Jet001 [13]
3 years ago
5

A, B, C, D, and E are collinear points. AE = 34, BD = 16, and AB = BC =CD. What is the length of CE?

Mathematics
2 answers:
musickatia [10]3 years ago
6 0

Answer:

CE= 18

Step-by-step explanation:

A, B, C, D, and E are collinear points

AE=AB +BC+CD+DE

BD = 16, and AB = BC =CD

BD=BC+CD, BC =CD

So BD= BC+BC

16=2BC

BC= 8

AB = BC =CD

AB= 8, BC=8, CD= 8, E= 34

AE=AB +BC+CD+DE

34=8+8+8+DE

34=24+DE, subtract 24 from both sides

DE= 10

CE= CD+DE

CE= 8+10

CE= 18

Lesechka [4]3 years ago
5 0
18. Drawn out we know that bc and cd are equal. Bd is 16 so bc and cd are both 8.
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Tyler owned the librarian late fees he had to pay a 3% fee on the cost of the books, which was $35.62. What was he charged for a
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$1.07 for the late fees.

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4 0
2 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).

(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.

7 0
3 years ago
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