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4 years ago

Please help will mark brainliest

Mathematics

1 answer:

Dmitry [639]4 years ago

3 0

Answer:

b.) 2376

Step-by-step explanation:

you will use the formula S.A.= 2lw+2lh+2hw

Hope this helps! Pls mark brainliest

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Which line pictured has a slope undefined?

Paraphin [41]

Slope A is undefined

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3 years ago

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The answer would be 250

sineoko [7]

Answer:

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Step-by-step explanation:

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3 years ago

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Write an equation of a line that passes through the two given points. Your answer should be written in slope-intercept form.

Vladimir79 [104]

Answer:

y = x

Step-by-step explanation:

Find the slope using rise over run (y2 - y1) / (x2 - x1)

Plug in the points:

(y2 - y1) / (x2 - x1)

(-8 - 0) / (-8 - 0)

-8 / -8

= 1

So, the slope is 1.

Plug in the slope and a point into y = mx + b and solve for b:

y = mx + b

-8 = 1(-8) + b

-8 = -8 + b

0 = b

Plug in the slope and y intercept into y = mx + b

y = mx + b

y = 1x + 0

y = x

So, the equation of the line is y = x

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3 years ago

A college student is interested in investigating the TV-watching habits of her classmates and surveys 20 people on the number of

Alla [95]

Answer:

80% confidence interval of the true average number of hours of TV watched per week is [8.28 hours, 11.02 hours].

Step-by-step explanation:

We are given that a college student is interested in investigating the TV-watching habits of her classmates and surveys 20 people on the number of hours they watch per week. The results are provided below;

Hours of TV per week (X): 6, 14, 13, 6, 16, 10, 19, 4, 5, 5, 18, 8, 7, 14, 8, 8, 9, 12, 6, 5.

Firstly, the Pivotal quantity for 80% confidence interval for the true average is given by;

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

where, $\bar X$ = sample mean number of hours of TV watched per week = $\frac{\sum X}{n}$ = 9.65

s = sample standard deviation = $\sqrt{\frac{\sum (X -\bar X)^{2} }{n-1} }$ = 4.61

n = sample of people = 20

$\mu$ = true average number of hours of TV watched per week

Here for constructing 80% confidence interval we have used One-sample t-test statistics as we don't know about population standard deviation.

So, 80% confidence interval for the true average, $\mu$ is ;

P(-1.33 < $t_1_9$ < 1.33) = 0.80 {As the critical value of t at 19 degrees of

freedom are -1.33 & 1.33 with P = 10%}

P(-1.33 < $\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }$ < 1.33) = 0.80

P( $-1.33 \times {\frac{s}{\sqrt{n} } }$ < ${\bar X-\mu}$ < $1.33 \times {\frac{s}{\sqrt{n} } }$ ) = 0.80

P( $\bar X-1.33 \times {\frac{s}{\sqrt{n} } }$ < $\mu$ < $\bar X+1.33 \times {\frac{s}{\sqrt{n} } }$ ) = 0.80

80% confidence interval for $\mu$ = [ $\bar X-1.33 \times {\frac{s}{\sqrt{n} } }$ , $\bar X+1.33 \times {\frac{s}{\sqrt{n} } }$ ]

= [ $9.65-1.33 \times {\frac{4.61}{\sqrt{20} } }$ , $9.65+1.33 \times {\frac{4.61}{\sqrt{20} } }$ ]

= [8.28 hours, 11.02 hours]

Therefore, 80% confidence interval of the true average number of hours of TV watched per week is [8.28 hours, 11.02 hours].

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3 years ago

4 cm 3 cm x cm area = 19 cm2

statuscvo [17]

Answer:

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3 years ago

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