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3 years ago

11

Which algebraic expression represents “Marcus practiced the trumpet twice as long this week”?2 + p2pp minus 2StartFraction p Ove

r 2 EndFraction

2 answers:

Tanya [424]3 years ago

6 0

Answer:

B) 2p

Step-by-step explanation:

muminat3 years ago

4 0

Answer:

B-2p

Step-by-step explanation:

Just took the quiz and got 100% good luck! And hope this helps!

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lianna [129]

Vanessa can make 5 portions

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2 years ago

Jaime makes the following claim.

Sphinxa [80]

Jaime is incorrect, the angle does not depend on the radius of the circles.

<h3>Is Jaime correct?</h3>

Remember that an angle that defines an arc on a circle, does not depend on the radius of the circle.

So, if we have an angle with a measure of π/3 radians in a circle with a radius of 3 inches and an angle with a measure of π/3 radians in a circle with a radius of 6 inches, these two angles are exactly the same thing.

The radius of the circle only has an impact on the length of the arc defined by the angle.

So Jaime is clearly incorrect.

If you want to learn more about angles:

brainly.com/question/17972372

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7 0

2 years ago

1 2 3 4 5 6 7 8 9 10 TIME REMAINING 52:51 Use the data set to determine which statements are correct. Check all that apply. 35,

Nady [450]

Answer:

The true statements are:

The median is 38.5.
There is an outlier.
The lower quartile is 29.
The upper quartile is 70.
The interquartile range is 41.

Step-by-step explanation:

The given data are: 35, 41, 18, 75, 36, 21, 62, 29, 154, 70

Make the data in order from the least to the greatest

18 , 21 , 29 , 35 , <u>36 , 41</u> , 62 , 70 , 75 , 154

So, the median of the data = (36 + 41)/2 = 38.5

We can deduce that The number 154 represents an outlier because the difference between 154 and 75 is much greater than the difference between the other data

If we divided the data to two groups , they will be:

(18 , 21 , <u>29 </u>, 35 , 36) , (41 , 62 , <u>70 </u>, 75 , 154)

So, The lower quartile Q1 is the median of the first group = 29

And the upper quartile Q2 is the median of the second group = 70

And the interquartile range = Q2 - Q1 = 70 - 29 = 41

So, the true statements are:

The median is 38.5
There is an outlier
The lower quartile is 29
The upper quartile is 70
The interquartile range is 41.

7 0

3 years ago

What is 28 divided by 3.2 ?? Please include steps :(

expeople1 [14]

thats the answer just look at the picture :)

for the work

8 0

3 years ago

An article contained the following observations on degree of polymerization for paper specimens for which viscosity times concen

devlian [24]

Answer:

(a) A 95% confidence interval for the population mean is [433.36 , 448.64].

(b) A 95% upper confidence bound for the population mean is 448.64.

Step-by-step explanation:

We are given that article contained the following observations on degrees of polymerization for paper specimens for which viscosity times concentration fell in a certain middle range:

420, 425, 427, 427, 432, 433, 434, 437, 439, 446, 447, 448, 453, 454, 465, 469.

Firstly, the pivotal quantity for finding the confidence interval for the population mean is given by;

P.Q. = $\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }$ ~ $t_n_-_1$

where, $\bar X$ = sample mean = $\frac{\sum X}{n}$ = 441

s = sample standard deviation = $\sqrt{\frac{\sum (X-\bar X)^{2} }{n-1} }$ = 14.34

n = sample size = 16

$\mu$ = population mean

<em>Here for constructing a 95% confidence interval we have used One-sample t-test statistics as we don't know about population standard deviation.</em>

<em />

<u>So, 95% confidence interval for the population mean, </u> $\mu$ <u> is ;</u>

P(-2.131 < $t_1_5$ < 2.131) = 0.95 {As the critical value of t at 15 degrees of

freedom are -2.131 & 2.131 with P = 2.5%}

P(-2.131 < $\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }$ < 2.131) = 0.95

P( $-2.131 \times {\frac{s}{\sqrt{n} } }$ < ${\bar X-\mu}$ < $2.131 \times {\frac{s}{\sqrt{n} } }$ ) = 0.95

P( $\bar X-2.131 \times {\frac{s}{\sqrt{n} } }$ < $\mu$ < $\bar X+2.131 \times {\frac{s}{\sqrt{n} } }$ ) = 0.95

<u>95% confidence interval for</u> $\mu$ = [ $\bar X -2.131 \times {\frac{s}{\sqrt{n} } }$ , $\bar X +2.131 \times {\frac{s}{\sqrt{n} } }$ ]

= [ $441-2.131 \times {\frac{14.34}{\sqrt{16} } }$ , $441+2.131 \times {\frac{14.34}{\sqrt{16} } }$ ]

= [433.36 , 448.64]

(a) Therefore, a 95% confidence interval for the population mean is [433.36 , 448.64].

The interpretation of the above interval is that we are 95% confident that the population mean will lie between 433.36 and 448.64.

(b) A 95% upper confidence bound for the population mean is 448.64 which means that we are 95% confident that the population mean will not be more than 448.64.

6 0

3 years ago