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Elanso [62]
3 years ago
12

Claire says the area of the square face is 8 m2. Is she correct? No, the face’s dimensions are 4 m by 7 m, so the area is 14 m2.

No, the face’s dimensions are 4 m by 4 m, so the area is 16 m2. No, the face’s dimensions are 4 m by 7 m, so the area is 28 m2. Yes, the face’s dimensions are 4 m by 4 m, so the area is 8 m2.
Mathematics
2 answers:
Len [333]3 years ago
7 0

Answer:

its B:No, the face’s dimensions are 4 m by 4 m, so the area is 16 m2.

motikmotik3 years ago
6 0

Answer:

its B:

Step-by-step explanation:

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The weights of steers in a herd are distributed normally. The variance is 40,00040,000 and the mean steer weight is 1400lbs1400
Alex777 [14]

Answer:

P(1739

(Correct to 4 decimal places)

Step-by-step explanation:

The probability of a continuous normal variable X found in a particular interval [a, b] is the area under the curve bounded by x=a and x=b and is given by:

P(a

where

f(X)=\frac{1}{\sigma \sqrt{2 \pi} }e^{-\frac{1}{2} (\frac{x-u}{ \sigma} )^2}



f(X)=\frac{1}{200 \sqrt{2 \pi} }e^{-\frac{1}{2} (\frac{x-1400}{ 200} )^2}\\=\frac{1}{200 \sqrt{2 \pi} }e^{- \frac{(x-1400)^2}{ 80000}\\

P(1739

P(1739

Using a calculator,

P(1739

5 0
3 years ago
Read 2 more answers
Find the appropriate rejection regions for the large-sample test statistic z in these cases. (Round your answers to two decimal
Usimov [2.4K]

Answer:

a) We have that the significance is given by \alpha =0.01 and we know that we have a right tailed test.

So for this case we need to look in the normal standard dsitribution a critical value that accumulates 1% of the area on the right and 99% of the area on the left. This value can be founded with the following excel code:

"=NORM.INV(1-0.01,0,1)"

And we got for this case z_{crit}=2.33

So then the rejection region would be z>2.33

b) We have that the significance is given by \alpha =0.05, \alpha/2 =0.025 and we know that we have a two tailed test.

So for this case we need to look in the normal standard dsitribution a critical value that accumulates 2.5% of the area on the right and 97.5% of the area on the left. This value can be founded with the following excel code:

"=NORM.INV(1-0.025,0,1)"

And we got for this case z_{crit}=\pm 1.96

So then the rejection region would be z>1.96 \cup z

Step-by-step explanation:

Part a

We have that the significance is given by \alpha =0.01 and we know that we have a right tailed test.

So for this case we need to look in the normal standard dsitribution a critical value that accumulates 1% of the area on the right and 99% of the area on the left. This value can be founded with the following excel code:

"=NORM.INV(1-0.01,0,1)"

And we got for this case z_{crit}=2.33

So then the rejection region would be z>2.33

Part b

We have that the significance is given by \alpha =0.05, \alpha/2 =0.025 and we know that we have a two tailed test.

So for this case we need to look in the normal standard dsitribution a critical value that accumulates 2.5% of the area on the right and 97.5% of the area on the left. This value can be founded with the following excel code:

"=NORM.INV(1-0.025,0,1)"

And we got for this case z_{crit}=\pm 1.96

So then the rejection region would be z>1.96 \cup z

7 0
3 years ago
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Hoochie [10]

Answer:

its 3hdnndndjsjshnanajzh z

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3 years ago
The equation y = 4x represents the relationship between time, x, and distance traveled, y. Which table represents this
Kruka [31]

Answer:

um

Step-by-step explanation:

um why did you spell a lot ????

4 0
3 years ago
Write the equation for the line that goes through point (1, 1) with slope m = -9
DENIUS [597]

1) Since we have the slope, and one point we can find that linear equation using this point-slope formula. Plugging them we have:

\begin{gathered} (y-y_0)=m(x-x_0) \\ y-1=-9(x-1) \\ y-1=-9x+9 \\ y=-9x+9+1 \\ y=-9x+10 \end{gathered}

2) Note that we had to add 1 to both sides, to keep the "y "on the left.

3) Therefore, y=-9x+10 is the answer.

7 0
11 months ago
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