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2 years ago

Solve the equation. S^2+ 7s = - 20

Mathematics

1 answer:

nataly862011 [7]2 years ago

8 0

this is called the quadratic equation.

Either use quadratic formula to find S or The PSF

Product, sum, factor box.

Quadratic formula is : (-b±√(b²-4ac))/(2a) .

we'll use the quadratic formula because psf ain't possible.

-7± √(49-80)/2

(-7±-5.57)/2

s= 0.715 or 6.285

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Help pls its just math :( Ik it sucks I hate math too, pls help tho

Umnica [9.8K]

Answer:

x = - 1

Step-by-step explanation:

Given

- 4(3x - 7) = 40 ( divide both sides by - 4 )

3x - 7 = - 10 ( add 7 to both sides )

3x = - 3 ( divide both sides by 3 )

x = - 1

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3 years ago

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If you split a right angle with a ray going through the middle of the right angle, the two angles formed will be ______. Select

Tom [10]

Answer:

Complementary

Step-by-step explanation:

To be complementary, your two angles need to add up to 90 degrees

a right triangle is already 90 degrees so if you cut it through the middle they'd form complementary angles

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3 years ago

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Suppose that a basketball player can score on a particular shot with probability .3. Use the central limit theorem to find the a

Rom4ik [11]

Answer:

(a) The probability that the number of successes is at most 5 is 0.1379.

(b) The probability that the number of successes is at most 5 is 0.1379.

(d) The probability that the number of successes is at most 11 is 0.9357.

→ All the exact probabilities are more than the approximated probability.

Step-by-step explanation:

Let S = a basketball player scores a shot.

The probability that a basketball player scores a shot is, P (S) = p = 0.30.

The number of sample selected is, n = 25.

The random variable $S\sim Bin(25,0.30)$

According to the central limit theorem if the sample taken from an unknown population is large then the sampling distribution of the sample proportion ( $\hat p$ ) follows a normal distribution.

The mean of the the sampling distribution of the sample proportion is: $E(\hat p)=p=0.30$

The standard deviation of the the sampling distribution of the sample proportion is:

$SD(\hat p)=\sqrt{\frac{ p(1- p)}{n} }=\sqrt{\frac{ 0.30(1-0.30)}{25} }=0.092$

(a)

Compute the probability that the number of successes is at most 5 as follows:

The probability of 5 successes is: $p=\frac{5}{25} =0.20$

$P(\hat p\leq 0.20)=P(\frac{\hat p-E(\hat p)}{SD(\hat p)}\leq \frac{0.20-0.30}{0.092} )\\=P(Z\leq -1.087)\\=1-P(Z$

**Use the standard normal table for probability.

Thus, the probability that the number of successes is at most 5 is 0.1379.

The exact probability that the number of successes is at most 5 is:

$P(S\leq 5)={25\choose 5}(0.30)^{5}91-0.30)^{25-5}=0.1935$

The exact probability is more than the approximated probability.

(b)

Compute the probability that the number of successes is at most 7 as follows:

The probability of 5 successes is: $p=\frac{7}{25} =0.28$

$P(\hat p\leq 0.28)=P(\frac{\hat p-E(\hat p)}{SD(\hat p)}\leq \frac{0.28-0.30}{0.092} )\\=P(Z\leq -0.2174)\\=1-P(Z$

**Use the standard normal table for probability.

Thus, the probability that the number of successes is at most 7 is 0.4129.

The exact probability that the number of successes is at most 7 is:

$P(S\leq 57)={25\choose 7}(0.30)^{7}91-0.30)^{25-7}=0.5118$

The exact probability is more than the approximated probability.

(c)

Compute the probability that the number of successes is at most 9 as follows:

The probability of 5 successes is: $p=\frac{9}{25} =0.36$

$P(\hat p\leq 0.36)=P(\frac{\hat p-E(\hat p)}{SD(\hat p)}\leq \frac{0.36-0.30}{0.092} )\\=P(Z\leq 0.6522)\\=0.7422$

**Use the standard normal table for probability.

Thus, the probability that the number of successes is at most 9 is 0.7422.

The exact probability that the number of successes is at most 9 is:

$P(S\leq 9)={25\choose 9}(0.30)^{9}91-0.30)^{25-9}=0.8106$

The exact probability is more than the approximated probability.

(d)

Compute the probability that the number of successes is at most 11 as follows:

The probability of 5 successes is: $p=\frac{11}{25} =0.44$

$P(\hat p\leq 0.44)=P(\frac{\hat p-E(\hat p)}{SD(\hat p)}\leq \frac{0.44-0.30}{0.092} )\\=P(Z\leq 1.522)\\=0.9357$

**Use the standard normal table for probability.

Thus, the probability that the number of successes is at most 11 is 0.9357.

The exact probability that the number of successes is at most 11 is:

$P(S\leq 11)={25\choose 11}(0.30)^{11}91-0.30)^{25-11}=0.9558$

The exact probability is more than the approximated probability.

6 0

3 years ago

You and your friend mix water and cirtic acid. You add 3 cups of citrtic acid for every 16 cups of water Your friend adds 2 cups

Vlad [161]

Answer:

A's mixture is more acidic.

Step-by-step explanation:

Given:

Two friends mix water and citric acid.

One (Let A) add 3 cups of citric acid for every 16 cups of water and another (Let B)adds 2 cups of citric acid for every 12 cups of water.

Now, we have to find that whose mixture is more acidic means:-

Solution:

A add 3 cups of citric acid for every 16 cups of water, the ratio = $\frac{3}{16}$

B adds 2 cups of citric acid for every 12 cups of water = $\frac{2}{12}$

Now, to find that who mix more citric acid in the mixture, firstly we will equalize the denominator of both fraction by taking LCM so that can be compare easily.

LCM of 16 and 12:- 48

16 - 16, 32, 48, 64, 80, 96....

12 - 12, 24, 36, 48, 60, 72.......

For friend A - $\frac{3\times3}{16\times3} =\frac{9}{48}$

For friend B - $\frac{2\times4}{12\times4} =\frac{8}{48}$

We found that in terms of 48 cups of water, friend A mixes 9 cups of acid while friend B mixes 8 cups of acid, means A is adding more acid than B and hence friend A's mixture is more acidic.

4 0

3 years ago

If arc VY = 132° and ∠YUV = 41°, what is the measure of arc XV?

Tom [10]

Answer:

Step-by-step explanation:

Use this equation

$\frac{b - a}{2} = angle \: yuv$