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3 years ago

Please assist meNO LINKS!!

Mathematics

1 answer:

Nana76 [90]3 years ago

4 0

Answer:

See below:

Step-by-step explanation:

Hello! My name is Galaxy and I will be helping you today, I hope you are having a nice day!

We can solve this in two steps, Comprehension and Solving. I'll go ahead and start with Comprehension.

If you have any questions feel free to ask away!!

Comprehension

We know according to the laws of geometry that all angles in a triangle add up to 180 degrees.

We also know that in an isosceles angle, the base angles are equation to each other.

Now that we know what we need to know, we can setup an equation.

$2x-6+(2x)=180$

We can do this because first of all, we know that 2x-6 is one of the angles and as per the base angles of an isosceles triangle we know that both base angles are x, therefore we can add 2x to get 180 degrees.

I'll start solving now.

Solving

We can solve this by using the equation we made above and solving it with algebra.

$2x-6+2x=180\\4x-6=180\\4x=186\\x=46.5$

We know that x is equal to 46.5 degrees. We can check that by inputting it into the equation.

$2(46.5)-6+2(46.5)=180?\\4(46.5)=186?\\46.5=46.5$

We've proven that our answer is correct by double checking,

Therefore the answer is 46.5!

Cheers!

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3 years ago

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15x^2 + 41x + 28 <br><br>STEP BY STEP ANSWER PLSSS

maria [59]

Answer:

(3x+4)(5x+7)

Step-by-step explanation:

15x^2 +41x+28

Factor the expression by grouping. First, the expression needs to be rewritten as 15x^2 +ax+bx+28. To find a and b, set up a system to be solved.

a+b=41

ab=15×28=420

Since ab is positive, a and b have the same sign. Since a+b is positive, a and b are both positive. List all such integer pairs that give product 420.

1,420

2,210

3,140

4,105

5,84

6,70

7,60

10,42

12,35

14,30

15,28

20,21

Calculate the sum for each pair.

1+420=421

2+210=212

3+140=143

4+105=109

5+84=89

6+70=76

7+60=67

10+42=52

12+35=47

14+30=44

15+28=43

20+21=41

The solution is the pair that gives sum 41.

a=20

b=21

Rewrite 15x^2 +41x+28 as (15x^2 +20x)+(21x+28).

(15x^2 +20x)+(21x+28)

Factor out 5x in the first and 7 in the second group.

5x(3x+4)+7(3x+4)

Factor out common term 3x+4 by using distributive property.

(3x+4)(5x+7)

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A = πab/4 solve for b

Studentka2010 [4]

Answer:

4A/ (πa) = b

Step-by-step explanation:

A = πab/4

Multiply each side by 4

4A = 4πab/4

4A = πab

Divide each side by πa

4A/(πa) = πab/ πa

4A/ (πa) = b

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3 years ago

The scores of students on the ACT college entrance exam in a recent year had the normal distribution with mean  =18.6 and stand

Maurinko [17]

Answer:

a) 33% probability that a single student randomly chosen from all those taking the test scores 21 or higher.

b) 0.39% probability that the mean score for 76 students randomly selected from all who took the test nationally is 20.4 or higher

Step-by-step explanation:

To solve this question, we need 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 $\sigma$ , the sample means with size n of at least 30 can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$

In this problem, we have that:

$\mu = 18.6, \sigma = 5.9$

a) What is the probability that a single student randomly chosen from all those taking the test scores 21 or higher?

This is 1 subtracted by the pvalue of Z when X = 21. So

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

$Z = \frac{21 - 18.6}{5.4}$

$Z = 0.44$

$Z = 0.44$ has a pvalue of 0.67

1 - 0.67 = 0.33

33% probability that a single student randomly chosen from all those taking the test scores 21 or higher.

b) The average score of the 76 students at Northside High who took the test was x =20.4. What is the probability that the mean score for 76 students randomly selected from all who took the test nationally is 20.4 or higher?

Now we have $n = 76, s = \frac{5.9}{\sqrt{76}} = 0.6768$