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VARVARA [1.3K]
3 years ago
12

What is the measure of angle MPJ?

Mathematics
1 answer:
jenyasd209 [6]3 years ago
7 0

where is the rest of the information?

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7x + 3 (5x + 9) = 247
Airida [17]
7x + 15x + 27 = 247
22x = 247 -27
22x =225
X= 10.22
8 0
4 years ago
4/18 is closest to: 1/2 or 1 whole
Anika [276]

Answer:

1/2, since 1 whole is 18/18 and 1/2 is equal to 9/18

Step-by-step explanation:

3 0
3 years ago
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A study investigated the effectiveness of meditation training in reducing trait anxiety. The study evaluated subjects trait anxi
Setler79 [48]

Answer:

The 99% confidence interval for the population mean reduction in anxiety was (1.2, 8.6).

Step-by-step explanation:

We have the standard deviation for the sample, which means that the t-distribution is used to solve this question.

The first step to solve this problem is finding how many degrees of freedom, we have. This is the sample size subtracted by 1. So

df = 27 - 1 = 26

99% confidence interval

Now, we have to find a value of T, which is found looking at the t table, with 26 degrees of freedom(y-axis) and a confidence level of 1 - \frac{1 - 0.99}{2} = 0.995. So we have T = 2.7787.

The margin of error is:

M = T\frac{s}{\sqrt{n}} = 2.7787\frac{6.9}{\sqrt{27}} = 3.7

In which s is the standard deviation of the sample and n is the size of the sample.

The lower end of the interval is the sample mean subtracted by M. So it is 4.9 - 3.7 = 1.2.

The upper end of the interval is the sample mean added to M. So it is 4.9 + 3.7 = 8.6.

The 99% confidence interval for the population mean reduction in anxiety was (1.2, 8.6).

6 0
3 years ago
What are the greatest common factors of 51 and 85
rusak2 [61]
Greatest\ common\ factor\ is\ the\ highest\ number\ by\ which\ 51\ and\ 85\\
can\ be\ divided\\\\ 
51:51\\\\
85:5\\
17:17\\\\GCF=1\\\\Greatest\ common\ factor\ of\ 51\ and\ 85\ is\ equal\ to\ 1.

6 0
3 years ago
A car purchased for $24,000 is expected to lose value, or depreciate, at a rate of 8% per year. This situation can be modeled by
algol [13]

\bf \qquad \textit{Amount for Exponential Decay} \\\\ A=P(1 - r)^t\qquad \begin{cases} A=\textit{accumulated amount}\\ P=\textit{initial amount}\dotfill &24000\\ r=rate\to 8\%\to \frac{8}{100}\dotfill &0.08\\ t=\textit{elapsed time}\dotfill &t\\ \end{cases} \\\\[-0.35em] ~\dotfill\\\\ \stackrel{\textit{current value}}{\stackrel{A}{\stackrel{\downarrow }{P(t)}}}=24000(1-0.08)^{\stackrel{\stackrel{\textit{years}}{\downarrow }}{t}}

5 0
3 years ago
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