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mrs_skeptik [129]
4 years ago
12

The width of a rectangle, in feet, is represented by (3x-1.5). The length of the rectangle, in feet, is represented by (1.25x+3)

. Find the perimeter of the rectangle.​
Mathematics
1 answer:
Tanya [424]4 years ago
5 0

Answer:

P =  8.5 x + 3  ft

Step-by-step explanation:

To find the perimeter, we use the formula

P =2(l+w) where l is the length and w is the width

P =2 (1.25x+3 + 3x-1.5)

Combine like terms

P = 2( 4.25x +1.5)

Distribute the 2

P =  8.5 x + 3  ft

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2x + 9 = 5x

5x = 2x + 9

Subtract sides 2x

- 2x + 5x = - 2x  + 2x + 9

3x = 9

Divide sides by 3

\frac{3x}{3}  =  \frac{9}{3}  \\

x = 3

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CHECK :

2x + 9 = 5x

2(3) + 9 = 5(3)

6 + 9 = 15

15 = 15

So the value is correct.

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8 0
3 years ago
Find the slop-intercept form of the line whose slope is 5 and that passes through the point (-6,7).
Korvikt [17]
You know that the slope, m=5. You have a point (6,7) as (x,y). The slope-intercept form is y=mx+b. Plug in the number for x,y, and m, and then you would be able to find b using the slope-intercept form. Then rewrite the equation as y=5x+b, which you just found b.
(Here's the first step. 7=5(-6)+b). Find b, and then write the slope-intercept form as y=5x+b. 
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Homework due at 12:00 I need HELP. Please!!!
vichka [17]
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4 0
3 years ago
According to a survey, high school girls average 100 text messages daily (The Boston Globe, April 21, 2010). Assume the populati
Ghella [55]

Answer:

Step-by-step explanation:

Hello!

The variable of interest is:

X: number of daily text messages a high school girl sends.

This variable has a population standard deviation of 20 text messages.

A sample of 50 high school girls is taken.

The is no information about the variable distribution, but since the sample is large enough, n ≥ 30, you can apply the Central Limit Theorem and approximate the distribution of the sample mean to normal:

X[bar]≈N(μ;δ²/n)

This way you can use an approximation of the standard normal to calculate the asked probabilities of the sample mean of daily text messages of high school girls:

Z=(X[bar]-μ)/(δ/√n)≈ N(0;1)

a.

P(X[bar]<95) = P(Z<(95-100)/(20/√50))= P(Z<-1.77)= 0.03836

b.

P(95≤X[bar]≤105)= P(X[bar]≤105)-P(X[bar]≤95)

P(Z≤(105-100)/(20/√50))-P(Z≤(95-100)/(20/√50))= P(Z≤1.77)-P(Z≤-1.77)= 0.96164-0.03836= 0.92328

I hope you have a SUPER day!

3 0
3 years ago
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