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Mathematics

1 answer:

Tom [10]2 years ago

4 0

Since the image has the same orientation as the original, it was not rotated or reflected. The only viable (and correct) choice is
.. D translation 4 units up.

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We really don't need to see your keyboard. It is appropriate to use a screen shot program or edit your picture, or both.

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Checks are stale dated when

lina2011 [118]

My guess is before the current date but not sure hope that helped and sorry I could not do more and always good luck

4 0

2 years ago

Read 2 more answers

△ABCis reflected to form △A′B′C′. The vertices of △ABC are A(3, 1), B(1, 5), and C(6, 9). The vertices of △A′B′C′ are A′(−1,

Blababa [14]

Answer: D) reflection across y = -x

Explanation:

When we reflect over y = x, we basically swap x and y. So for instance, the point (3,1) becomes (1,3).

When reflecting over y = -x, we will do the same thing but we'll make each coordinate swap in sign from positive to negative (or vice versa). The rule for reflecting over y = -x is $(x,y) \to (-y,-x)$

So if we apply that rule to point A(3,1) then it becomes A ' (-1, -3).

Similarly, B(1,5) moves to B ' (-5, -1)

Finally, C(6,9) becomes C ' (-9, -6)

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2 years ago

The accompanying summary data on total cholesterol level (mmol/l) was obtained from a sample of Asian postmenopausal women who w

dimulka [17.4K]

Answer:

99% CI:

$-4.637\leq\mu_v-\mu_o\leq 3.737$

Step-by-step explanation:

We have to calculate a 99%CI for the difference of means for the vegan and the omnivore.

First, we have to estimate the standard deviation

$s_{Md}=\sqrt{\frac{2MSE}{n_h}}$

The MSE can be calculated as

$MSE=\frac{(SSE_1+SSE_2)}{df} =\frac{(n_1s_1)^2+(n_2*s_2)^2}{n_1+n_2-2}\\\\MSE=\frac{(85*1.05)^2+(96*1.20)^2}{85+96-2}=\frac{7965.56+13272.04}{179} =118.65$

The weighted sample size nh can be calculated as

$n_h=\frac{2}{1/n_1+1/n_2}=\frac{2}{1/85+1/96}=\frac{2}{0.0222} =90.16$

The standard deviation then becomes

$s_{Md}=\sqrt{\frac{2MSE}{n_h}}=\sqrt{\frac{2*118.65}{90.16}}=\sqrt{2.632} =1.623$

The z-value for a 99% confidence interval is z=2.58.

The difference between means is

$\Delta M=M_v-M_o=5.10-5.55=-0.45$

Then the confidence interval can be constructed as

$\Delta M-z*s_{Md}\leq\mu_v-\mu_o\leq \Delta M+z*s_{Md}\\\\ -0.45-2.58*1.623\leq\mu_v-\mu_o\leq -0.450+2.58*1.623\\\\-0.450-4.187\leq\mu_v-\mu_o\leq-0.450+4.187\\\\-4.637\leq\mu_v-\mu_o\leq 3.737$