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3 years ago

How do I finish this problem? I keep getting stumped here

Mathematics

2 answers:

prisoha [69]3 years ago

5 0

I believe you would Add like terms so 4x^2-4x-3

castortr0y [4]3 years ago

5 0

1.)Break the expression into groups
(4x^2+2x)+(-6x-3)
2.) Factor out
2x(2x+1)-3(2x+1)
3.)Factor out common term
(2x+1)(2x-3)

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What is 32gal/min=qt/h

bagirrra123 [75]

480 quarts and hour is the answer

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3 years ago

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inessss [21]

the answer is 16x^2-4

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3 years ago

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WILL MARK BRAINLIEST!!!NEED HELP ASAP!! Carrie will spin the arrow on the spinner four times. What is the probability that the a

sladkih [1.3K]

The probability that the arrow will land on any of them once is 1/4.

The probability that the arrow will land on any of them twice is 1/4 * 1/4. This is because the probability of Event A and Event B is P(B) * P(A).

Based on this:

$P(a\cap b\cap c\cap d) = \frac{1}{4}\times \frac{1}{4}\times\frac{1}{4}\times\frac{1}{4}\\\\=\frac{1}{4\times4\times4\times 4}\\=\frac{1}{256}$

1/256

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3 years ago

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Let us suppose we have data on the absorbency of paper towels that were produced by two different manufacturing processes. From

maksim [4K]

Answer:

The 95% CI for the difference of means is:

$-155.45 \leq \mu_1-\mu_2 \leq -44.55$

Step-by-step explanation:

The question is incomplete:

"Find a 95% confidence interval on the difference of the towels mean absorbency produced by the two processes. Assumed that the standard deviations are estimated from the data. Round to two decimals places."

Process 1:

- Sample size: 10

- Mean: 200

- S.D.: 15

Process 2:

- Sample size: 4

- Mean: 300

- S.D.: 50

The difference of the sample means is:

$M_d=M_1-M_2=200-300=-100$

The standard deviation can be estimated as:

$\sigma_d=\sqrt{\frac{\sigma_1^2}{n_1}+\frac{\sigma_2^2}{n_2}}\\\\\sigma_d=\sqrt{\frac{15^2}{10}+\frac{50^2}{4}} =\sqrt{22.5+625}=\sqrt{647.5}=25.45$

The degrees of freedom are:

$df=n_1+n_2-2=10+4-2=12$

The t-value for a 95% confidence interval and 12 degrees of freedom is t=±2.179.

Then, the confidence interval can be written as:

$M_d-t\cdot \sigma_d\leq \mu_1-\mu_2 \leq M_d+t\cdot \sigma_d\\\\-100-2.179*25.45\leq \mu_1-\mu_2 \leq -100+2.179*25.45\\\\-100-55.45 \leq \mu_1-\mu_2 \leq -100+55.45\\\\ -155.45 \leq \mu_1-\mu_2 \leq -44.55$

8 0

3 years ago

Hodel just got her driver's license. Her parents offer her two options. 1. They'll buy her a used car that will cost her $0.50 p

Romashka-Z-Leto [24]

Answer:

The second option will cost her less than the first one.

Step-by-step explanation:

In order to solve this problem we will create two functions to represent the cost of the car in function of the miles drove by her.

For the first option we have:

$cost_1(x) = 0.5*x$

For the second option we have:

$cost_2(x) = 5000 + 0.1*x$

Since she intends to drive it for 10,000 miles per year for 6 years, then the total mileage she intends to drive her car is 60,000 miles. Applying this to the formula of each car and we have:

$cost_1(60000) = 0.5*60000 = 30000$

$cost_2(60000) = 5000 + 60000*0.1 = 11000$

The second option will cost her less than the first one.

7 0

3 years ago