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jeka94
3 years ago
15

A student completed the following problem. Justify each step: 8x + (14 + x) = 59 8x + (x + 14) = 59 (8x + x) + 14 = 59 9x + 14 =

59 9x + 14 - 14 = 59 - 14 9x = 45 x = 5
Mathematics
1 answer:
viva [34]3 years ago
7 0

Answer:

Step-by-step explanation:

Mutiply 8x×14+8x×xlike that way solve

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Please help me look at the attached picture
Lady bird [3.3K]

Answer:

Q(-10) = 299

Step-by-step explanation:

Q(x) = 3x^{2} - 1Q(-10) = 3(-10)^{2} - 1

Q(-10) = 3(100) - 1  (-10 to the second power is 100 (negative times a negative equals a positive)

Q(-10) = 300 - 1  3 times 100 equal 300

Q(-10) = 299

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4 years ago
What is three and 2 fourths and minus two and three fourths
netineya [11]

Answer:

The answer is 0.75

Hope this helps!

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2 years ago
What do the following two equations represent?
Dima020 [189]

Answer:

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3 years ago
The article "Students Increasingly Turn to Credit Cards" (San Luis Obispo Tribune, July 21, 2006) reported that 37% of college f
Sloan [31]

Answer:

Step-by-step explanation:

Hello!

There are two variables of interest:

X₁: number of college freshmen that carry a credit card balance.

n₁= 1000

p'₁= 0.37

X₂: number of college seniors that carry a credit card balance.

n₂= 1000

p'₂= 0.48

a. You need to construct a 90% CI for the proportion of freshmen  who carry a credit card balance.

The formula for the interval is:

p'₁±Z_{1-\alpha /2}*\sqrt{\frac{p'_1(1-p'_1)}{n_1} }

Z_{1-\alpha /2}= Z_{0.95}= 1.648

0.37±1.648*\sqrt{\frac{0.37*0.63}{1000} }

0.37±1.648*0.015

[0.35;0.39]

With a confidence level of 90%, you'd expect that the interval [0.35;0.39] contains the proportion of college freshmen students that carry a credit card balance.

b. In this item, you have to estimate the proportion of senior students that carry a credit card balance. Since we work with the standard normal approximation and the same confidence level, the Z value is the same: 1.648

The formula for this interval is

p'₂±Z_{1-\alpha /2}*\sqrt{\frac{p'_2(1-p'_2)}{n_2} }

0.48±1.648* \sqrt{\frac{0.48*0.52}{1000} }

0.48±1.648*0.016

[0.45;0.51]

With a confidence level of 90%, you'd expect that the interval [0.45;0.51] contains the proportion of college seniors that carry a credit card balance.

c. The difference between the width two 90% confidence intervals is given by the standard deviation of each sample.

Freshmen: \sqrt{\frac{p'_1(1-p'_1)}{n_1} } = \sqrt{\frac{0.37*0.63}{1000} } = 0.01527 = 0.015

Seniors: \sqrt{\frac{p'_2(1-p'_2)}{n_2} } = \sqrt{\frac{0.48*0.52}{1000} }= 0.01579 = 0.016

The interval corresponding to the senior students has a greater standard deviation than the interval corresponding to the freshmen students, that is why the amplitude of its interval is greater.

8 0
4 years ago
Need help failing in math
ASHA 777 [7]

Answer:

If im not incorrect the answer should be 5

Step-by-step explanation:

yes it is because look at it like this f(10)=50 so if u think what is 50 divided by 10 you will equal up to 5

4 0
3 years ago
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