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Alexus [3.1K]
3 years ago
8

Could anyone help me find the surface area of this? I got 340 and I don't think it's right.

Mathematics
1 answer:
Dmitry_Shevchenko [17]3 years ago
3 0

Answer: I got 200

Step-by-step explanation: 14+14+5+5=38*2=76 Thats the front and back and then The sides are 6+6+5+5=22*2=44 then the top and bottom are 14 +14+6+6=40*2=80 then add that all up and 44+80+76=200 (Sorry if its a bit confusing to understand)

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Find the area of the sector is formed by NMP round your answer to the nearest hundredth of a square centimeter
Shalnov [3]

Answer:

44.68 square cm

Step-by-step explanation:

To find the area of the whole circle you use the formula pie x r^2

then multiply that area by 40/360 because that is the area of the sector

4 0
2 years ago
−3(3h−1)≤−7h−5<br> kkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkk
Lady bird [3.3K]

Answer:

h<=4

Step-by-step explanation:

-9h+3<=-7h-5

-9h+7h<=-5-3

-2h<=-8

h<=4

3 0
2 years ago
CAN SOMEONE PLEASE JUST ANSWER THIS ASAP FOR BRAINLIEST!!!
Julli [10]

Maybe the answer is -4 ....? I just can't find any more logic to this problem.

5 0
3 years ago
A photo is 4 inches by 6 inches. What is the area of the photo?
kolezko [41]
4 * 6 = 24
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Hope this helped :)
4 0
3 years ago
An article in Fire Technology investigated two different foam-expanding agents that can be used in the nozzles of firefighting s
UNO [17]

Answer:

Step-by-step explanation:

Hello!

The objective of this experiment is to test if two different foam-expanding agents have the same foam expansion capacity

Sample 1 (aqueous film forming foam)

n₁= 5

X[bar]₁= 4.7

S₁= 0.6

Sample 2 (alcohol-type concentrates )

n₂= 5

X[bar]₂= 6.8

S₂= 0.8

Both variables have a normal distribution and σ₁²= σ₂²= σ²= ?

The statistic to use to make the estimation and the hypothesis test is the t-statistic for independent samples.:

t= \frac{(X[bar]_1 - X[bar]_2) - (mu_1 - mu_2)}{Sa*\sqrt{\frac{1}{n_1} + \frac{1}{n_2 } } }

a) 95% CI

(X[bar]_1 - X[bar]_2) ± t_{n_1 + n_2 - 2}*Sa* \sqrt{\frac{1}{n_1}+\frac{1}{n_2} }

Sa²= \frac{(n_1-1)S_1^2 + (n_2-1)S_2^2}{n_1 + n_2 - 2}= \frac{(5-1)0.36 + (5-1)0.64}{5 + 5 - 2}= 0.5

Sa= 0.707ç

t_{n_1 + n_2 -2: 1 - \alpha /2} = t_{8; 0.975} = 2.306

(4.7-6.9) ± 2.306* (0.707\sqrt{\frac{1}{5}+\frac{1}{5} })

[-4.78; 0.38]

With a 95% confidence level you expect that the interval [-4.78; 0.38] will contain the population mean of the expansion capacity of both agents.

b.

The hypothesis is:

H₀: μ₁ - μ₂= 0

H₁: μ₁ - μ₂≠ 0

α: 0.05

The interval contains the cero, so the decision is to reject the null hypothesis.

<u>Complete question</u>

a. Find a 95% confidence interval on the difference in mean foam expansion of these two agents.

b. Based on the confidence interval, is there evidence to support the claim that there is no difference in mean foam expansion of these two agents?

8 0
3 years ago
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