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Grace [21]
3 years ago
8

Find the Quotient 8372divided by 5

Mathematics
1 answer:
igomit [66]3 years ago
8 0

Answer:

The correct answer would be 1674.4

Step-by-step explanation:

1. Write the problem in long division format.

2. Divide 8 by 5 to get 1.

3. Multiply the quotient digit (1) by the divisor 5.

4. Subtract 5 from 8 to get 3.

5. Bring down the next number of the dividend.

6. Divide 33 by 5 to get 6.

7. Multiply the quotient digit (6) by the divisor 5.

8. Subtract 30 from 33.

9. Bring down the next number of the dividend.

10. Divide 37 by 5 to get 7

11. Multiply the quotient digit (7) by the divisor 5 .

12. Subtract 35 from 37.

13. Bring down the next number of the dividend.

14. Divide 22 by 5 to get 4

15. Multiply the quotient digit (4) by the divisor 5

16. Subtract 20 from 22.

The solution for the Long Division of 8372/5 is 1674 with a remainder of 2 or as 1674.4

(I am sorry I am not able to show the long division, but I hope this helps)

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What is the probability of of selecting a lemon piece out of 6
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Answer:

I'm assuming it's 1 lemon piece among 5 other pieces, if so then 1 out of 6.

Step-by-step explanation:

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2 years ago
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Solve the inequality (picture)
drek231 [11]

Answer:

3rd choice

Step-by-step explanation:

Add 6 to both sides to get 5x >= 15, or x >=3. This is the 3rd choice.

8 0
3 years ago
Group Work 2.6 Newton's Law of Cooling states that an object cools at a rate proportional to the difference between the temperat
Airida [17]

Answer:

ΔT = ΔT0 e^-K T

As I understand Newton's Law of Cooling

ΔT at any time is the difference between the temperature and the surroundings

Originally       ΔT0 = 95 - 22   difference between 95 and room temperature

65 - 22 = 33 = 73 e^-KT      where t is time to cool to 65 deg

ln (33/73) = -KT       K = .794 / 5 = .159    where 5 is time to cool to 65 deg

40 - 22 = 73 e^-.159 T      where t is time to cool to 40 deg

18 = 73 e^-.159 T

ln (18 / 73) = -.159 T

T = 8.8 min

It would take 8.8 min for the object to cool to 40 deg C

Suppose the object cooled from 95 to 90 deg, then

ln 68 / 73 = -.159 T     and T = .45 min

7 0
2 years ago
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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
What is 190 x 173 idk what it is and i have a bunch to do
Anna007 [38]
32,870 is the answer. Hope this helps!
6 0
3 years ago
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