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Pavel [41]
2 years ago
11

Andrea received a $40 gift card to use at a town for a cost her eight dollars per visit to swim a function relating the value of

the gift card, v, to the number of visits, n, is v(n) = 40-8n
Mathematics
1 answer:
BigorU [14]2 years ago
5 0

Answer:

n = v/8 = 40/8 = 5

Step-by-step explanation:

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A 273,000-gallon shipment of fuel is 2.3 percent
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Answer:

Amount of fuel antifreeze = 6,279 gallon

Step-by-step explanation:

Given:

Amount of fuel =  273,000-gallon

Antifreeze = 2.3 %

Find:

Amount of fuel antifreeze

Computation:

Amount of fuel antifreeze = Amount of fuel × Antifreeze

Amount of fuel antifreeze = 273,000 × 2.3 %

Amount of fuel antifreeze = 6,279 gallon

8 0
3 years ago
3. At the beginning of the day, a water tank contained 526.8 gallons of water. During the day,
Andru [333]

Answer:

208.85

Step-by-step explanation:

It is 208.85 because when you have .8 + .05 =.85

5 0
2 years ago
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What is 4/5 of 20.5?
ira [324]
16.4
20.5/5=4.1
4.1*4=16.4 
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7 0
3 years ago
(9 1/3 + 4 2/3) +7 - (-12)= <br><br> A: 2 and 5/8<br> B: -13 and 13/16<br> C: 14<br> D: 4
SSSSS [86.1K]

Answer:

BIL is the answer to that one day

7 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).

(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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