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juin [17]
2 years ago
11

X + y = 92 and y = 3x - 4

Mathematics
1 answer:
Advocard [28]2 years ago
6 0

∑ Hey, eduardolozanosalgado ⊃

Answer:

x=24 , y=68

Step-by-step explanation:

<u><em>Given:</em></u>

<u><em></em></u>\mathrm{\begin{bmatrix}x+y=92\\ y=3x-4\end{bmatrix}}

<u><em>Solution:</em></u>

<em>Substitute: </em>\mathrm{y = 3x-4}

\begin{bmatrix}x+3x-4=92\end{bmatrix}

Isolate x for 4x - 4= 92 : x = 24

Substitute : x = 24

y=3\cdot \:24-4

y=64

Hence, the solutions are : x=24,\:y=68

<u><em>xcookiex12</em></u>

<u><em></em></u>

<em>8/19/2022</em>

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Roderick is counting the change on his dresser. he has twice as many pennies as he does dimes. he has twice the amount of dimes
melamori03 [73]
P = 2d
d = 2n.....n = 1/2d
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since dimes is mentioned in relation to  quarters, pennies, and the nickels, u would use d as ur variable

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8 0
3 years ago
A random sample of 49 lunch customers was taken at a restaurant. The average amount of time the customers in the sample stayed i
Hunter-Best [27]

Answer:

a)  σ/√n= 1.43 min

c) Margin of error 2.8028min

d) [30.1972; 35.8028]min

e) n=62 customers

Step-by-step explanation:

Hello!

The variable of interest is

X: Time a customer stays at a restaurant. (min)

A sample of 49 lunch customers was taken at a restaurant obtaining

X[bar]= 33 mi

The population standard deviation is known to be δ= 10min

a) and b)

There is no information about the distribution of the population, but we know that if the sample is large enough, n≥30, we can apply the central limit theorem and approximate the distribution of the sample mean to normal:

X[bar]≈N(μ;σ²/n)

Where μ is the population mean and σ²/n is the population variance of the sampling distribution.

The standard deviation of the mean is the square root of its variance:

√(σ²/n)= σ/√n= 10/√49= 10/7= 1.428≅ 1.43min

c)

The CI for the population mean has the general structure "Point estimator" ± "Margin of error"

Considering that we approximated the sampling distribution to normal and the standard deviation is known, the statistic to use to estimate the population mean is Z= (X[bar]-μ)/(σ/√n)≈N(0;1)

The formula for the interval is:

[X[bar]±Z_{1-\alpha /2}*(σ/√n)]

The margin of error of the 95% interval is:

Z_{1-\alpha /2}= Z_{1-0.025}= Z_{0.975}= 1.96

d= Z_{1-\alpha /2}*(σ/√n)= 1.96* 1.43= 2.8028

d)

[X[bar]±Z_{1-\alpha /2}*(σ/√n)]

[33±2.8028]

[30.1972; 35.8028]min

Using a confidence level of 95% you'd expect that the interval [30.1972; 35.8028]min contains the true average of time the customers spend at the restaurant.

e)

Considering the margin of error d=2.5min and the confidence level 95% you have to calculate the corresponding sample size to estimate the population mean. To do so you have to clear the value of n from the expression:

d= Z_{1-\alpha /2}*(σ/√n)

\frac{d}{Z_{1-\alpha /2}}= σ/√n

√n*(\frac{d}{Z_{1-\alpha /2}})= σ

√n= σ* (\frac{Z_{1-\alpha /2}}{d})

n=( σ* (\frac{Z_{1-\alpha /2}}{d}))²

n= (10*\frac{1.96}{2.5})²= 61.47≅ 62 customers

I hope this helps!

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Answer:

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Step-by-step explanation:

if you subtract 55 from 90 and divide it by 5 then you get your answer, which is C:7

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