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3 years ago

Plzzz help plzzz I need help on both of them

Mathematics

1 answer:

Paul [167]3 years ago

5 0

The answer to the first problem is G.

Since there are 3 boys and 9 girls who are left-handed, that makes 12 people in total in sixth grade who are left-handed. The total number of sixth graders at the middle school is 150 since there are 90 girls and 60 boys in that grade. Divide 150 from 12 to get 0.08. You change it to percentage form by multiplying 0.08 by 100 to get 8%.

The answer for the second problem is C. You change the percentage (in this case it is 76%) to decimal form by dividing by 100 to get 0.76. You then multiply 0.76 with 8,950 to get 6,802.

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Solve and graph –7 < w – 4 < 2

Nikitich [7]

Answer:

solution: -3<w<6

interval notation: (-3,6)

Step-by-step explanation:

The coordinates are (-3,6) all you have to do is put them on a graph! Hope this helps!

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3 years ago

Find the surface area of the prism.

anygoal [31]

Answer:

220 I'm pretty sure because 6×5 is 30 times 2 is 60. Then do 6×10 equals 60. Then do 5 times 10 times 2 to get 100. Then you add it all up to get 220

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3 years ago

Read 2 more answers

Ms. Bergstedt asked two students to come up with two different sequences. Jay created this sequence: The first term is 4 and eac

Roman55 [17]

Well, create an equation.

y=6x+4 and y=5x+8

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This means it's the fourth term and the number is 28.

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3 years ago

How many planes can be drawn through any three noncollinear points?

RideAnS [48]

Answer:

1 plane

I hope this helps!

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4 years ago

An educational psychologist wishes to know the mean number of words a third grader can read per minute. She wants to make an est

zhannawk [14.2K]

Answer:

99% confidence interval for the mean number of words a third grader can read per minute is [24.7 , 25.1].

Step-by-step explanation:

We are given that an educational psychologist wishes to know the mean number of words a third grader can read per minute. She wants to make an estimate at the 99% level of confidence. For a sample of 4657 third graders, the mean words per minute read was 24.9. Assume a population standard deviation of 5.7.

So, the pivotal quantity for 99% confidence interval for the mean number of words a third grader can read per minute is given by;

P.Q. = $\frac{\bar X - \mu}{\frac{\sigma}{\sqrt{n} } }$ ~ N(0,1)

where, $\bar X$ = sample mean words per minute read = 24.9

$\sigma$ = population standard deviation = 5.7

n = sample of third graders = 4657

$\mu$ = population mean

So, 99% confidence interval for the population mean, $\mu$ is ;

P(-2.5758 < N(0,1) < 2.5758) = 0.99

P(-2.5758 < $\frac{\bar X - \mu}{\frac{\sigma}{\sqrt{n} } }$ < 2.5758) = 0.99

P( $-2.5758 \times {\frac{\sigma}{\sqrt{n} } }$ < ${\bar X - \mu}$ < $2.5758 \times {\frac{\sigma}{\sqrt{n} } }$ ) = 0.99

P( $\bar X-2.5758 \times {\frac{\sigma}{\sqrt{n} } }$ < $\mu$ < $\bar X+2.5758 \times {\frac{\sigma}{\sqrt{n} } }$ ) = 0.99

<u>99% confidence interval for </u> $\mu$ = [ $\bar X-2.5758 \times {\frac{\sigma}{\sqrt{n} } }$ , $\bar X+2.5758 \times {\frac{\sigma}{\sqrt{n} } }$ ]

= [ $24.9-2.5758 \times {\frac{5.7}{\sqrt{4657} } }$ , $24.9+2.5758 \times {\frac{5.7}{\sqrt{4657} } }$ ]

= [24.7 , 25.1]

Therefore, 99% confidence interval for the mean number of words a third grader can read per minute is [24.7 , 25.1].

6 0

3 years ago