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3 years ago

Find the sum of the terms.

Mathematics

1 answer:

kow [346]3 years ago

5 0

Answer:

Step-by-step explanation:

Given

$\frac{x}{x^2+3x+2}$ + $\frac{3}{x+1}$ ← factor the denominator of first fraction

= $\frac{x}{(x+1)(x+2)}$ + $\frac{3}{x+1}$

Multiply the numerator/ denominator of second fraction by (x + 2)

= $\frac{x}{(x+1)(x+2)}$ + $\frac{3(x+2)}{(x+1)(x+2)}$

The denominators are now common, so add the numerators leaving the denominator, that is

= $\frac{x+3x+6}{(x+1)(x+2)}$

= $\frac{4x+6}{(x+1)(x+2)}$ → with numerator 4x + 6 → C

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The mean points obtained in an aptitude examination is 159 points with a standard deviation of 13 points. What is the probabilit

Korolek [52]

Answer:

0.4514 = 45.14% probability that the mean of the sample would differ from the population mean by less than 1 point if 60 exams are sampled

Step-by-step explanation:

To solve this question, we have to understand the normal probability distribution and the central limit theorem.

Normal probability distribution:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

Central limit theorem:

The Central Limit Theorem estabilishes that, for a random variable X, with mean $\mu$ and standard deviation $\sigma$ , a large sample size can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$

In this problem, we have that:

$\mu = 159, \sigma = 13, n = 60, s = \frac{13}{\sqrt{60}} = 1.68$

What is the probability that the mean of the sample would differ from the population mean by less than 1 point if 60 exams are sampled?

This is the pvalue of Z when X = 159+1 = 160 subtracted by the pvalue of Z when X = 159-1 = 158. So

X = 160

$Z = \frac{X - \mu}{\sigma}$

By the Central Limit Theorem

$Z = \frac{X - \mu}{s}$

$Z = \frac{160 - 159}{1.68}$

$Z = 0.6$

$Z = 0.6$ has a pvalue of 0.7257

X = 150

$Z = \frac{X - \mu}{s}$

$Z = \frac{158 - 159}{1.68}$

$Z = -0.6$

$Z = -0.6$ has a pvalue of 0.2743

0.7257 - 0.2743 = 0.4514

0.4514 = 45.14% probability that the mean of the sample would differ from the population mean by less than 1 point if 60 exams are sampled

7 0

3 years ago

Please solve i will give brainiest 100 point question ****** do the whole page please need to pass or i will fail its my final t

dmitriy555 [2]

Answer:

1. Find the difference between the areas.

Area of the small rectangle: $(x+2)(x+7)=x^2+7x+2x+14=x^2+9x+14$

Area of the big rectangle: $(x+9)(x+11)=x^2+11x+9x+99=x^2+20x+99$

The difference is: $11x+85$

$( x^2+20x+99)- (x^2+9x+14)=x^2+20x+99-x^2-9x-14=11x+85$

You can solve this question just by looking at the graph.

a) The height is 4 meters.

$f(d)=h=-2d^2+7d+4$

To find the height of the bleachers, we should consider the moment before the shoot, when the distance is equal to 0.

$f(0)=h=-2(0)^2+7(0)+4$

$h=4$

The height is 4 meters.

b) 9 meters.

For $d=1$

$f(1)=h=-2(1)^2+7(1)+4$

$f(1)=h=-2+7+4$

$h=9$

b) The ball travels 4 meters.

But to calculate it, it is when $h=0$

$0=-2d^2+7d+4$

Using the quadratic formula:

$d=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

$d=\frac{-7 \pm \sqrt{7^2-4\left(-2\right)4}}{2\left(-2\right)}$

$d=\frac{-7\pm\sqrt{81}}{-4}$

$d=\frac{-7\pm9}{-4}$

It will give us to solutions, once it is a quadratic equation, but we are talking about a positive distance.

$d=-\frac{1}{2} \text{ or }d=4$

In this question, we have to find the area of the cylinder and the sphere.

From the information given, we have

a = 5mm and d = 5mm, therefore the radius is 2.5 mm.

The volume of a cylinder:

$V=\pi r^2h$

$V=\pi (2.5)^2 \cdot 5$

$V=31.25 \pi$

$V_{c} \approx 98.17 \text{ m}^3$

The volume of the sphere:

$V=\frac{4}{3} \pi r^2$

$V_{s} \approx 65.4 \text{ m}^3$

The volume of the capsule is approximately $163.57 \text{ m}^3$

3 0

3 years ago

Multiply and and simplify : (a+b)^2 Pls help easy

Lena [83]

Answer:a^2+2ab+b^2

Step-by-step explanation:

First write it out in extended form, (a+b)(a+b) then multiply each term by the other in each parenthesis, such as a*a+b*b+a*b+a*b

Then, once you have done that, simplify it. you will then get the answer a^2+2ab+b^2

3 0

3 years ago

Please help.

m_a_m_a [10]

Answer:

26.2

Step-by-step explanation:

PQ/QR = 8/16 = 0.5

tan(26.6) = 0.5

7 0

3 years ago

How many 1/2s are in 4?

WARRIOR [948]

Answer:

Step-by-step explanation:

5 0

2 years ago