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3 years ago

5a²-4ab-b²/b²+7ab+10a²

1 answer:

6 0

15a^2+3ab-1, I would explain my answer, however, I would get deleted for copying something else which is ridiculous.<span />

You might be interested in

A right triangle has legs of lengths 6 and 15 inches. Find the length of the hypotenuse (nearest tenth).

zubka84 [21]

Answer:

The length of the hypotenuse is 16.2 inches.

Step-by-step explanation:

$c^{2} = a^{2} + b^{2}$

$c^{2} = 6^{2} + 15^{2}$

$c^{2} = 36 + 225$

$c^{2} = 261$

$\sqrt{c^{2} } = \sqrt{261}$

$c = 16.2$

8 0

3 years ago

Sila jogged 12 kilometers today and wants to know how far she traveled in feet. What error did Sila make while converting from k

ozzi

Answer:

umm i'm not rly sure what's going on here...but here's how i'd do it

Step-by-step explanation:

3280.84x12=39370.08 which is about 3,9370 feet.

to double check let's do an algebraic expression

1/3280.84=x/12 cross multiply

x(1)=3280.84(12) simplify

x=39370.08

hope that helped.

8 0

3 years ago

**Two trains are traveling at constant speeds on different tracks. What is the speed of Train B? *

Eduardwww [97]

Answer:

um what?

Step-by-step explanation:

4 0

3 years ago

WILL BE MARKED BRAINLEST ANSWER. There are 1,062 souvenir paperweights that need to be packed in boxes. Each box will hold 12 pa

Brrunno [24]

I'd say 88.5 because 12 divided by 1062 is 88.5 therefor thats how many boxes youd need.

4 0

3 years ago

2. When a large truckload of mangoes arrives at a packing plant, a random sample of 150 is selected and examined for

kirza4 [7]

a) The 90% confidence interval of the percentage of all mangoes on the truck that fail to meet the standards is: (7.55%, 12.45%).

b) The margin of error is: 2.45%.

c) The 90% confidence is the level of confidence that the true population percentage is in the interval.

d) The needed sample size is: 271.

<h3>What is a confidence interval of proportions?</h3>

A confidence interval of proportions has the bounds given by the rule presented as follows:

$\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}$

The margin of error is given by:

$M = z\sqrt{\frac{\pi(1-\pi)}{n}}$

The variables are listed as follows:

$\pi$ is the sample proportion, which is also the estimate of the parameter.

z is the critical value.

n is the sample size.

The confidence level is of 90%, hence the critical value z is the value of Z that has a p-value of $\frac{1+0.90}{2} = 0.95$ , so the critical value is z = 1.645.

The sample size and the estimate are given as follows:

$n = 150, \pi = \frac{15}{150} = 0.1$

The margin of error is of:

$M = z\sqrt{\frac{0.1(0.9)}{150}} = 0.0245 = 2.45\%$

The interval is given by the estimate plus/minus the margin of error, hence:

The lower bound is: 10 - 2.45 = 7.55%.

The upper bound is: 10 + 2.45 = 12.45%.

For a margin of error of 3% = 0.03, the needed sample size is obtained as follows:

$M = z\sqrt{\frac{\pi(1-\pi)}{n}}$

$0.03 = 1.645\sqrt{\frac{0.1(0.9)}{n}}$

$0.03\sqrt{n} = 1.645\sqrt{0.1(0.9)}$

$\sqrt{n} = \frac{1.645\sqrt{0.1(0.9)}}{0.03}$

$(\sqrt{n}})^2 = \left(\frac{1.645\sqrt{0.1(0.9)}}{0.03}\right)^2$

n = 271 (rounded up).

More can be learned about the z-distribution at brainly.com/question/25890103

#SPJ1

3 0

2 years ago