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Delvig [45]
3 years ago
13

Find the difference in simplest form:

Mathematics
2 answers:
zvonat [6]3 years ago
8 0
The correct answer is D
Hope this helped
Andrew [12]3 years ago
4 0

Answer:

D. is the answer I got i think it is right but IDK

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The area of the large rectangle is twice the area of the small rectangle. What is the value of x?Two rectangles are shown. The l
kupik [55]

Answer:

B. 8 cm.

Step-by-step explanation:

The area of a rectangle is the length*width.

The area of the larger rectangle is 4*12=48.

The area of the smaller rectangle is 3*x=3x.

If the larger rectangle's area is twice that of the smaller rectangle, then we can say 48=2*3x.

Let's solve for x

48=2*3x

48=6x

x=8

3 0
3 years ago
Amelie is shopping for children's books and puzzle books. She wants to purchase at least 2 more children's books than puzzle boo
oksano4ka [1.4K]
X = 9
y = 6

Step by step:

Divide 15 by 2: 15 ÷ 2 = 7.5
Subtract 0.5 from 7.5 (this will later be added back): 7.5 - 0.5 = 7
Subtract 1 from 7 (she wants to buy at least 2 more children books): 7 - 1 = 6
Subtract 6 from 15 (6 being the number of puzzle books, 15 being the total number of items): 15 - 6 = 9 (number of children books)
6 0
3 years ago
Help me plz thank you
nevsk [136]

Answer:

1) Distance=Speed*time

90=(x+1)*(x)+(2x+5)(x-1)

90=x^2+x+2x^2-2x+5x-5

3x^2+4x-95=0

2) 3x^2+4x-95=0. Using quadratic formula, we get

x=(-4±sqrt(16-4*3*(-95)<u>)</u>)/6, x=5 or - 19/5 but since x also represents time, it can't be negative.

3) Total time take she took for the journey is x+x-1=2x-1=2*5-1=9 hours

4 0
3 years ago
A plane flies 1400 kilometres in 2 hours 30 minutes
Sloan [31]
560 kilometers flied in each hour
8 0
3 years ago
Jensen Tire &amp; Auto is in the process of deciding whether to purchase a maintenance contract for its new computer wheel align
sasho [114]

Answer:

Step-by-step explanation:

Hello!

The given data corresponds to the variables

Y:  Annual Maintenance  Expense ($100s)

X: Weekly Usage  (hours)

n= 10

∑X= 253; ∑X²= 7347; \frac{}{X}= ∑X/n= 253/10= 25.3 Hours

∑Y= 346.50; ∑Y²= 13010.75; \frac{}{Y}= ∑Y/n= 346.50/10= 34.65 $100s

∑XY= 9668.5

a)

To estimate the slope and y-intercept you have to apply the following formulas:

b= \frac{sumXY-\frac{(sumX)(sumY)}{n} }{sumX^2-\frac{(sumX)^2}{n} } = \frac{9668.5-\frac{253*346.5}{10} }{7347-\frac{(253)^2}{10} }= 0.95

a= \frac{}{Y} -b\frac{}{X} = 34.65-0.95*25.3= 10.53

^Y= a + bX

^Y= 10.53 + 0.95X

b)

H₀: β = 0

H₁: β ≠ 0

α:0.05

F= \frac{MS_{Reg}}{MS_{Error}} ~~F_{Df_{Reg}; Df_{Error}}

F= 47.62

p-value: 0.0001

To decide using the p-value you have to compare it against the level of significance:

If p-value ≤ α, reject the null hypothesis.

If p-value > α, do not reject the null hypothesis.

The decision is to reject the null hypothesis.

At a 5% significance level you can conclude that the average annual maintenance expense of the computer wheel alignment and balancing machine is modified when the weekly usage increases one hour.

b= 0.95 $100s/hours is the variation of the estimated average annual maintenance expense of the computer wheel alignment and balancing machine is modified when the weekly usage increases one hour.

a= 10.53 $ 100s is the value of the average annual maintenance expense of the computer wheel alignment and balancing machine when the weekly usage is zero.

c)

The value that determines the % of the variability of the dependent variable that is explained by the response variable is the coefficient of determination. You can calculate it manually using the formula:

R^2 = \frac{b^2[sumX^2-\frac{(sumX)^2}{n} ]}{[sumY^2-\frac{(sumY)^2}{n} ]} = \frac{0.95^2[7347-\frac{(253)^2}{10} ]}{[13010.75-\frac{(346.50)^2}{10} ]} = 0.86

This means that 86% of the variability of the annual maintenance expense of the computer wheel alignment and balancing machine is explained by the weekly usage under the estimated model ^Y= 10.53 + 0.95X

d)

Without usage, you'd expect the annual maintenance expense to be $1053

If used 100 hours weekly the expected maintenance expense will be 10.53+0.95*100= 105.53 $100s⇒ $10553

If used 1000 hours weekly the expected maintenance expense will be $96053

It is recommendable to purchase the contract only if the weekly usage of the computer is greater than 100 hours weekly.

4 0
3 years ago
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