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vitfil [10]
2 years ago
5

ANSWER IS C

Mathematics
1 answer:
MA_775_DIABLO [31]2 years ago
8 0

Answer:

x <2

Step-by-step explanation:

2.5 – 1.2x < 6.5 – 3.2x

Add 3.2x to each side

2.5 – 1.2x+3.2x < 6.5 – 3.2x+3.2x

2.5 +2x < 6.5

Subtract 2.5 from each side

2.5+2x-2.5<6.5-2.5

2x<4

Divide by 2

2x/2 < 4/2

x <2

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Answer:

<u>The Answer in Exact Form:</u>

-3/22

<u>In Decimal Form:</u>

-0.136

Step-by-step explanation:

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A train drove the distance of 440 feet in 20 seconds. Select ALL the unit rates that are equivalent to the speed of the model ai
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Answer:

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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
1000 was invested at 5.5% interest, compounded annually. After sometime the amount had grown to 1550. How long was the money inv
scZoUnD [109]

compound interest equation for annually compounded

A=P(1+r)^t

A=final amount

P=principal

r=rate in decimal

t=time in years


given that

A=1550

P=1000

r=5.5%=0.055

find t


1550=1000(1+0.055)^t

divide both sides by 1000

1.55=1.055^t

take ln of both sides

ln(1.55)=ln(1.055^t)

use ln rule ln(a^b)=b(ln(a))

ln(1.55)=t(ln(1.055))

divide both sides by ln(1.055)

\frac{ln(1.55)}{ln(1.055)}=t

using a calculator, we get that t=8.18544 yrs

so about 8.2yrs

4 0
3 years ago
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