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Alik [6]
2 years ago
6

If you can, please tell how you solved

Mathematics
1 answer:
gizmo_the_mogwai [7]2 years ago
8 0

Answer:

plzz mark me brainlieat if it helps you

x+67 + x+127 = 180

or, 2x + 194 = 180

2x = - 14

: . x = -7

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The answer is y=2x+5
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3 years ago
Lorenzo is planning a 285-mile trip. His car’s EPA rating is 18 mpg. The number of gallons of gas that Lorenzo will need for the
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It would require 15.8333333 to make that trip.
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At the end of each year, Carl and Linda Munson will deposit $2,100 into a 401k retirement account.
Vanyuwa [196]

Answer:

you get the answer yet im trying to figure it out to

Step-by-step explanation:

6 0
3 years ago
Merina is scheduled to make two loan payments to Bradford in the amount of $1,000 each, two months and nine months from now. Mer
Svetach [21]

well, we're assuming all along that Merina owes Bradford $2000, because in the 1st scenario, she was going to pay twice $1000.

on the 2nd scenario, she'll be paying the same $2000 but split 7 months from now and then 7 months later, same 2000 bucks, at which point Bradford applied 8.5% interest.

using those assumptions, since the wording is not quite clear, we can say that Merina is simply paying 2000 bucks plus the 8.5%

\begin{array}{|c|ll} \cline{1-1} \textit{a\% of b}\\ \cline{1-1} \\ \left( \cfrac{a}{100} \right)\cdot b \\\\ \cline{1-1} \end{array}~\hspace{5em}\stackrel{\textit{8.5\% of 2000}}{\left( \cfrac{8.5}{100} \right)2000}\implies 170 \\\\[-0.35em] ~\dotfill\\\\ \cfrac{\stackrel{principal}{2000}~~ + ~~\stackrel{interest}{170}}{2}\implies \stackrel{\textit{two equal payments of}}{1085}

6 0
2 years ago
Evelyn wants to estimate the percentage of people who own a tablet computer she surveys 150 indvidals and finds that 120 own a t
igomit [66]

Answer:

The 99% confidence interval for the percentage of people who own a tablet computer is between 71.59% and 88.41%

Step-by-step explanation:

Confidence interval for the proportion of people who own a tablet:

In a sample with a number n of people surveyed with a probability of a success of \pi, and a confidence level of 1-\alpha, we have the following confidence interval of proportions.

\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}

In which

z is the zscore that has a pvalue of 1 - \frac{\alpha}{2}.

For this problem, we have that:

n = 150, \pi = \frac{120}{150} = 0.8

99% confidence level

So \alpha = 0.01, z is the value of Z that has a pvalue of 1 - \frac{0.01}{2} = 0.995, so Z = 2.576.

The lower limit of this interval is:

\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.8 - 2.575\sqrt{\frac{0.8*0.2}{150}} = 0.7159

The upper limit of this interval is:

\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.8 + 2.575\sqrt{\frac{0.8*0.2}{150}} = 0.8841

Percentage:

Multiply the proportion by 100.

0.7159*100 = 71.59%

0.8841*100 = 88.41%

The 99% confidence interval for the percentage of people who own a tablet computer is between 71.59% and 88.41%

8 0
3 years ago
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