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3 years ago

The figures shown are similar

1 answer:

8 0

Hi there!

Based on your options, I think we're going from the big figure to the small figure. 12/8 = 1 1/2

ANSWER:
D - 1.5

Hope this helps!! :)
If there's anything else that I can help you with, please let me know!

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Determine which postulate or theorem can be used to prove that

EleoNora [17]

Answer:

$\triangle ABC \cong \triangle DCB$ by AAS

Step-by-step explanation:

According to the following two triangles, $\triangle ABC$ and $\triangle DCB$ are congruent by Angle-Angle-Side (AAS), because there are two angles shown and share a side, which is in the middle between triangles.

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3 years ago

Please help me Evaluate -9 - (-4)

Y_Kistochka [10]

Answer:

Should be 13. Sorry if I’m incorrect!

Step-by-step explanation:

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3 years ago

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EVERYONE, PLEASE HELP I NEED THIS DONE

zimovet [89]

Answer:

$\frac{4}{5}$

Step-by-step explanation:

$\frac{32}{40}$

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40÷8=5

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1 year ago

Read 2 more answers

Boxes of raisins are labeled as containing 22 ounces. Following are the weights, in the ounces, of a sample of 12 boxes. It is r

ZanzabumX [31]

Answer:

A 90% confidence interval for the mean weight is [21.78 ounces, 21.98 ounces].

Step-by-step explanation:

We are given the weights, in the ounces, of a sample of 12 boxes below;

Weights (X): 21.88, 21.76, 22.14, 21.63, 21.81, 22.12, 21.97, 21.57, 21.75, 21.96, 22.20, 21.80.

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 weight = $\frac{\sum X}{n}$ = 21.88 ounces

s = sample standard deviation = $\sqrt{\frac{\sum (X-\bar X)^{2} }{n-1} }$ = 0.201 ounces

n = sample of boxes = 12

$\mu$ = population mean weight

Here for constructing a 90% confidence interval we have used a One-sample t-test statistics because we don't know about population standard deviation.

So, 90% confidence interval for the population mean, $\mu$ is ;

P(-1.796 < $t_1_1$ < 1.796) = 0.90 {As the critical value of t at 11 degrees of

freedom are -1.796 & 1.796 with P = 5%}

P(-1.796 < $\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }$ < 1.796) = 0.90

P( $-1.796 \times {\frac{s}{\sqrt{n} } }$ < ${\bar X-\mu}$ < $1.796 \times {\frac{s}{\sqrt{n} } }$ ) = 0.90

P( $\bar X-1.796 \times {\frac{s}{\sqrt{n} } }$ < $\mu$ < $\bar X+1.796 \times {\frac{s}{\sqrt{n} } }$ ) = 0.90

90% confidence interval for $\mu$ = [ $\bar X-1.796 \times {\frac{s}{\sqrt{n} } }$ , $\bar X+1.796 \times {\frac{s}{\sqrt{n} } }$ ]

= [ $21.88-1.796 \times {\frac{0.201}{\sqrt{12} } }$ , $21.88+1.796 \times {\frac{0.201}{\sqrt{12} } }$ ]

= [21.78, 21.98]

Therefore, a 90% confidence interval for the mean weight is [21.78 ounces, 21.98 ounces].

8 0

3 years ago

Sqrt(0.32)*(2sqrt(800)-3sqrt(1800))

Iteru [2.4K]

Answer:

paki ayos pohhh salamat or patingin Ng example para ma answer kupo salamat

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3 years ago