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bagirrra123 [75]
2 years ago
13

A local hamburger shop sold a combined total of 591 hamburgers and cheeseburgers on Tuesday. There were 59 fewer cheeseburgers s

old than hamburgers
On tuesday how much did they sell
Mathematics
2 answers:
hodyreva [135]2 years ago
6 0

Answer:

<u>On Tuesday</u>

591 burgers are sold.

Some are hamburgers and some are cheeseburgers.

59 fewer cheeseburgers sold than hamburgers.

Number of cheeseburgers --> x

Number of hamburgers --> y

x+y= 591 --> x = 591-y

x = y-59

so

591-y = y-59

-2y = -650

y = 325

so

x = 325-59

x =266

Papessa [141]2 years ago
4 0
591/2= 295.5-59=236.5

591-236.5=354.5

Cheeseburgers sold: 236.5
Hamburgers: 354.5

354.5+236.5= 591 (total sold)

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Step-by-step explanation:

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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.

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Sholpan [36]

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Answer:

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x is replaced by 2 so 2+6=8

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