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3 years ago

QUESTION BELOW PLS HELP

Mathematics

1 answer:

Sidana [21]3 years ago

3 0

Answer:

B. V-Vo/t

Step-by-step explanation:

All you have to do is redistribute the equation

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5x6 = (_X 6) + (2x_)

kati45 [8]

Answer:

(3 x 6) + (2 x 6)

Step-by-step explanation:

6 0

2 years ago

A businessman was comparing his company's profits for two consecutive months. In November, n, were 0.04 thousand dollars more th

Firdavs [7]

N and d are both numbers of thousands of dollars.
Thus, if n = 1, that means $1000.

Here n = d + 0.4.
Note that in C, n+0.04=n is completely wrong.
Similarly, in D, d = 0.04 = n is completely wrong.
The "combined profit amount" for Nov. and Dec. is n + d = 3.15.

Only A matches this info. Your answer is A.

5 0

3 years ago

Please help me (30 points) 1. A recipe calls for 2 1/3 cups of flour, 3/4 cups of white sugar, and 1/3 cups of brown sugar. The

Katena32 [7]

Lets find how much flour we need for just one serving.

2 2/3 ÷ 6

8/3 ÷ 6

8/3 * 1/6

8/18

4/9 cup of flour per serving

If we triple the recipe (6 * 3 = 18), we will have a total of 18 servings. Multiply the amount of flour we got for one serving to 18 servings.

4/9 * 18/1

72/9

8 cups of flour for 18 servings

Best of Luck!

4 0

3 years ago

Help please ASAP !! !!!!

goblinko [34]

Answer:

im gonna say 23 or 24 not 100% sure

Step-by-step explanation:

3 0

2 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.

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.

Complete question

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