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2 years ago

What is the solution to the system of equations graphed below? y = -4x +33 y = 5x-1

Mathematics

1 answer:

alex41 [277]2 years ago

7 0

It can be solved.

X = 34/9

Y= 161/9

Working in photo.

Please give brainliest.

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Which list of fractions is in order from smallest to largest? A. 1, 7/10, 3/10, 1/10 B. 3/10, 7/10, 1/10, 1 C. 1/10, 3/10, 7/10,

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Answer:

Step-by-step explanation:

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3 years ago

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Lily visits 100 patients in a month and records the age of each patient. The pie chart below shows the number of patients she ha

iogann1982 [59]

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3 years ago

An experiment was conducted as a one-way factorial design with K sample means, each based on n scores. What is the number of deg

nordsb [41]

Answer:

The mean squares has d.f (n-1)

Step-by-step explanation:

The total number of degrees of freedom is n-1 as there is only one restriction of computing the grand mean. The d.f for k samples is k-1 beacuase the mean of the sample means must equal the grand mean. Similarly , the d.f for within SS is n-k , due to the k restrictions of computing the k sample means. Hence we find that

Total df= Within df + Between df

n-1= (n-k)+(k-1)

Between SS has (k-1) d.f

Within SS has (n-k) d.f

These two quantities are known as mean squares and has d.f (n-1)

8 0

3 years ago

The length of a rectangular box is less than 2 inches more than the width. if the length is 7 inches, what is the area of the bo

kifflom [539]

They tell you that the length is 7 inches and that it is 2 inches more than the width therefore you can conclude that: length = 7 and the width = 5
you can plug the values given into the formula listed below:

A = lw
A = 7 x 5
A = 35

hope this helps :)

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2 years ago

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A small lawnmower company produced 1,500 lawnmowers in 2008. In an effort to determine how maintenance-free these units were, th

ikadub [295]

Answer:

The 95% confidence interval for the average number of years until the first major repair is (3.1, 3.5).

Step-by-step explanation:

The (1 - α)% confidence interval for the average using the finite correction factor is:

$CI=\bar x\pm z_{\alpha/2}\cdot\frac{\sigma}{\sqrt{n}}\cdot\sqrt{\frac{N-n}{N-1}}$

The information provided is:

$N=1500\\n=183\\\sigma=1.47\\\bar x=3.3$

The critical value of z for 95% confidence level is,

z = 1.96

Compute the 95% confidence interval for the average number of years until the first major repair as follows:

$CI=\bar x\pm z_{\alpha/2}\cdot\frac{\sigma}{\sqrt{n}}\cdot\sqrt{\frac{N-n}{N-1}}$

$=3.3\pm 1.96\times\frac{1.47}{\sqrt{183}}\times\sqrt{\frac{1500-183}{1500-1}}\\\\=3.3\pm 0.19964\\\\=(3.10036, 3.49964)\\\\\approx (3.1, 3.5)$

Thus, the 95% confidence interval for the average number of years until the first major repair is (3.1, 3.5).

7 0

3 years ago