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morpeh [17]
4 years ago
12

I WILL AWARD BRAINLIEST!

Mathematics
2 answers:
sammy [17]4 years ago
8 0
A)70 this is the answer
gavmur [86]4 years ago
7 0

Answer:

A. 70

Step-by-step explanation:

You can use the transversals and use vertical angles are congruent.

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How do I solve this step by step?
Mashcka [7]
First step : 2(2))+5\2 -2(2))+3\2 =9\2 -7\2 =2\2 =1
5 0
3 years ago
The distance between John's house and Albert’s house is 8 1/3 miles. A park is located on a straight path between the two houses
Murrr4er [49]

Answer:

The average mass of a group of children is 50 kilograms. Todd, who has a mass of 62 kilograms, then joins the group. This raises the average mass of the group to 52 kilograms. How many children were in the original group

Step-by-step explanation:

4 0
3 years ago
The following table shows the data collected from four random samples of 50 students from a small middle school regarding their
Sonbull [250]

Answer:

Answer: The reading that is preferred by the majority of students in the samples is: Novels.

Step-by-step explanation:

We are given table of data that represents the reading preferences of 50 students as:Sample# Novels Short Stories Classics ComicBooks Total 1 26 19 1 4 50 2 30 10 3 7 50 3 33 15 1 1 50 4 22 18 2 8 50Clearly from all the 4 samples we could observe that most students like: Novels.and then they love Short stories , after which they love to read Comic books and the least preferred readings is: Classics.

4 0
3 years ago
Determine the number of solutions for the system of equations y=2/3x+4 and y=2/3-x2
Phoenix [80]

2/3x+4=2/3-2x

2/3x-(-2x)

2/3x+2x

2 2/3x+4=2/3

2/3-4=-3 1/3

2 2/3x=-3 1/3

(3 1/3)/(2 2/3)=1 1/4

x = 1 1/4

y=2/3-2x

y=2/3-2(2/3)

y=2/3-4/3

y=-2/3

The answer is (1 1/4, -2/3)

4 0
4 years ago
A random sample of 300 Americans planning long summer vacations in 2009 revealed an average planned expenditure of $1,076. The e
Rufina [12.5K]

Answer:

(1024.69,\ 1127.31)

Step-by-step explanation:

We know that the sample size was:

n = 300

The average was:

{\displaystyle {\overline {x}}}=1,076

The standard deviation was:

\s = 345

The confidence level is

1-\alpha = 0.99

\alpha=1-0.99\\\alpha=0.01

The confidence interval for the mean is:

{\displaystyle{\overline {x}}} \± Z_{\frac{\alpha}{2}}*\frac{s}{\sqrt{n}}

Looking at the normal table we have to

Z_{\frac{\alpha}{2}}=Z_{\frac{0.01}{2}}=Z_{0.005}=2.576

Therefore the confidence interval for the mean is:

1,076\± 2.576*\frac{345}{\sqrt{300}}

1,076\± 51.31

(1024.69,\ 1127.31)

This means that <em>the mean planned spending of all Americans who take long summer vacations in 2009 is between $ 1024.69 and $ 1127.31</em>

3 0
4 years ago
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