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3 years ago

Write the integers in order from least to greatest

Mathematics

1 answer:

Charra [1.4K]3 years ago

8 0

#7) -7,-2,0,3,4
#8) -9,-6,-5,1,8

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4.9.18

Kruka [31]

Answer:

x= -5 or -1

Step-by-step explanation:

subst y = -x + 5 into y = x^2 + 5x + 10

-x +5 = x^2 + 5x +10

x^2 +6x +5 = 0

(x+5)(x+1)=0

x= -5 or -1

3 0

3 years ago

Read 2 more answers

This question says “use the given conditions to write an equation for the line in point-slope form and in slope intercept form”.

Ipatiy [6.2K]

Answer:

Step-by-step explanation:

y = -4x + 5.....slope here is -4. A parallel line will have the same slope.

y = mx + b

slope(m) = -4

(-6,-2)....x = -6 and y = -2

now we sub and find b, the y intercept

-2 = -4(-6) + b

-2 = 24 + b

-2 - 24 = b

-26 = b

so ur equation is : y = -4x - 26 <====

4 0

3 years ago

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Gabriella made 5% of her free throws over the season. If she shot 180 free throws, how

faust18 [17]

171. Have a nice day!

7 0

3 years ago

Read 2 more answers

3=x+9<br> PLEASE SHOW WORK. SOLVE AND CHECK.

Vika [28.1K]

Answer:

x = —6

Step-by-step explanation:

3 = x + 9

Collect like terms

3 — 9 = x

— 6 = x

x = —6

4 0

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

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

8 0

3 years ago