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elena55 [62]
3 years ago
11

5 years ago, Cheryl was d years old, Brandon is 2 years older than Cheryl.

Mathematics
1 answer:
lana [24]3 years ago
4 0

Answer: See explanation

Step-by-step explanation:

a. how old is Cheryl?

Cheryl's age = d + 5

b. how old is Brandon?

d + 5 + 2

= d + 7

c. what was the difference in their ages 5 years ago?

Cheryl age five years ago = d

Brandon's age five years ago = d + 2

Difference = d + 2 - d = 2 years

d. what is the sum of their ages now?

Cheryl's age = d + 5

Brandon age = d + 7

Sum = d + 5 + d + 7

= 2d + 12

e. what will the sum of their ages be two years from now?

Two years from now,

Cheryl's age = d + 5 + 2 = d + 7

Brandon age = d + 7 + 2 = d + 9

Sum = d + 7 + d + 9

= 2d + 16

f. what will the difference of their ages be two years from now

Two years from now,

Cheryl's age = d + 5 + 2 = d + 7

Brandon age = d + 7 + 2 = d + 9

Difference = Brandon age - Cheryl age

= (d + 9) - (d + 7)

= 2 years.

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Ab is 10. 8 units long if ABC is dilated by a scale factor of k equals 1.3 what is the length of a'b'
Marizza181 [45]
ANSWER


The length of

A'B'=14.04 \: units

EXPLANATION

The object length given to us is,

|AB|=10.8 \: units

The scale factor of the dilation is,

k = 1.3

The image length can be calculated using the formula

|A'B'| = k \times  |AB|

We substitute the given values into the formula to get,

|A'B'| = 1.3\times 10.8


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|A'B'| = 14.04 \: units
7 0
3 years ago
Solve the following equations<br><br> 12-(3x+13)=0
coldgirl [10]

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x=-1/3

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12-(3x+13)=0

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6 0
3 years ago
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Blizzard [7]

Answer:

t=\frac{26.8-29.9}{\frac{2.7}{\sqrt{9}}}=-3.44    

df=n-1=9-1=8  

p_v =P(t_{(8)}  

Since the p value is lower than the significance level we have enough evidence to reject the null hypothesis and is not enough evidence to conclude that the claim is true.

Step-by-step explanation:

Data given

\bar X=26.8 represent the sample mean

s=2.7 represent the sample standard deviation

n=9 sample size  

\mu_o =29.9 represent the value that we want to test

\alpha=0.05 represent the significance level for the hypothesis test.  

t would represent the statistic (variable of interest)  

p_v represent the p value for the test (variable of interest)  

System of hypothesis

We need to conduct a hypothesis in order to check if the true mean exceed 29.9 or no, the system of hypothesis would be:  

Null hypothesis:\mu \geq 29.9  

Alternative hypothesis:\mu < 29.9  

The statistic is given by:

t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}  (1)  

Calculate the statistic

We can replace in formula (1) the info given like this:  

t=\frac{26.8-29.9}{\frac{2.7}{\sqrt{9}}}=-3.44    

P-value

The degrees of freedom are given by:  

df=n-1=9-1=8  

Since is a one sided test the p value would be:  

p_v =P(t_{(8)}  

Conclusion  

Since the p value is lower than the significance level we have enough evidence to reject the null hypothesis and is not enough evidence to conclude that the claim is true.

6 0
4 years ago
3 + x - 7x = -15 + 3x
asambeis [7]
X=2
To check it you plug in 2 for x. Which then gives you:

3+(2-14)=-15+6 which then gives you:
3+-12=-15+6 which the gives you:
-9=-9

6 0
4 years ago
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