Find the area (pls help)

What are the solutions of the equation (x-3)2 + 2(x – 3) - 8= 0? Use u substitution to solve.

astraxan [27]

Answer:

x = - 1 and x = 5

Step-by-step explanation:

Given

(x - 3)² + 2(x - 3) - 8 = 0

Let u = x - 3, then

u² + 2u - 8 = 0 ← solve for u

(u + 4)(u - 2) = 0 ← in factored form

Equate each factor to zero and solve for u

u + 4 = 0 ⇒ u = - 4

u - 2 = 0 ⇒ u = 2

Convert back to x, that is

x - 3 = - 4 ( add 3 to both sides ) ⇒ x = - 1

x - 3 = 2 ( add 3 to both sides ) ⇒ x = 5

5 0

4 years ago

Read 2 more answers

Please help me, thank you very much

son4ous [18]

Y=2x-3
If u put in the values of x U will see that it equals the corresponding y value.

5 0

3 years ago

How many solutions does the system have

Delicious77 [7]

Answer:

There are infinate solutions.

Step-by-step explanation:

Plug (3x-8) in for y

5(3x-8)=15x-40

Simplify

15x-40=15x-40

4 0

3 years ago

Read 2 more answers

What are the terms in the expression 10p + 3q + 2?

Rufina [12.5K]

Answer:

Step-by-step explanation:

10p. 39 and z. the terms of 10p+3q+2 are lop, 39 ands In polynomials, each monomial is called the term of the.

5 0

3 years ago

In the following problem, check that it is appropriate to use the normal approximation to the binomial. Then use the normal dist

frosja888 [35]

Answer:

a) 0.9920 = 99.20% probability that 15 or more will live beyond their 90th birthday

b) 0.2946 = 29.46% probability that 30 or more will live beyond their 90th birthday

c) 0.6273 = 62.73% probability that between 25 and 35 will live beyond their 90th birthday

d) 0.0034 = 0.34% probability that more than 40 will live beyond their 90th birthday

Step-by-step explanation:

We solve this question using the normal approximation to the binomial distribution.

Binomial probability distribution

Probability of exactly x sucesses on n repeated trials, with p probability.

Can be approximated to a normal distribution, using the expected value and the standard deviation.

The expected value of the binomial distribution is:

$E(X) = np$

The standard deviation of the binomial distribution is:

$\sqrt{V(X)} = \sqrt{np(1-p)}$

Normal probability distribution

Problems of normally distributed distributions can be 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.

When we are approximating a binomial distribution to a normal one, we have that $\mu = E(X)$ , $\sigma = \sqrt{V(X)}$ .

In this problem, we have that:

Sample of 723, 3.7% will live past their 90th birthday.

This means that $n = 723, p = 0.037$ .

So for the approximation, we will have:

$\mu = E(X) = np = 723*0.037 = 26.751$

$\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{723*0.037*0.963} = 5.08$

(a) 15 or more will live beyond their 90th birthday

This is, using continuity correction, $P(X \geq 15 - 0.5) = P(X \geq 14.5)$ , which is 1 subtracted by the pvalue of Z when X = 14.5. So

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

$Z = \frac{14.5 - 26.751}{5.08}$

$Z = -2.41$

$Z = -2.41$ has a pvalue of 0.0080

1 - 0.0080 = 0.9920

0.9920 = 99.20% probability that 15 or more will live beyond their 90th birthday

(b) 30 or more will live beyond their 90th birthday

This is, using continuity correction, $P(X \geq 30 - 0.5) = P(X \geq 29.5)$ , which is 1 subtracted by the pvalue of Z when X = 29.5. So

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

$Z = \frac{29.5 - 26.751}{5.08}$

$Z = 0.54$

$Z = 0.54$ has a pvalue of 0.7054

1 - 0.7054 = 0.2946

0.2946 = 29.46% probability that 30 or more will live beyond their 90th birthday

(c) between 25 and 35 will live beyond their 90th birthday

This is, using continuity correction, $P(25 - 0.5 \leq X \leq 35 + 0.5) = P(X 24.5 \leq X \leq 35.5)$ , which is the pvalue of Z when X = 35.5 subtracted by the pvalue of Z when X = 24.5. So