Can u help me please

Factor completely.<br> 486 + 108z + 6x2 =

Firlakuza [10]

Answer:

6(x^2 +18z +81) . . . as written
6(x +9)^2 . . . . . . . . assuming a typo

Step-by-step explanation:

As the expression is written, the only factoring that can be done is to factor out the common numeric factor of 6:

= 6(x^2 +18z +81)

__

<em>Assuming a typo in the question</em>

If we assume all variables are supposed to be "x", then the quadratic term can be further factored as a square:

= 6(x +9)^2

_____

It is often handy to remember that the form of the square of a binomial is ...

(a +b)^2 = a^2 +2ab +b^2

Here, we recognize that 18 = 2√81, so the binomial is (x +√81) = (x +9).

7 0

3 years ago

I need the distance please

sasho [114]

Use this formula: $d= \sqrt{(x_{2}-x_{1})^2+(y_{2}-y_{1})^2 }$

Again :'(

5. d=10

6. d=4

7. d=4

5 0

4 years ago

12. A bucket contains 40 yellow golf balls and 30 white golf balls. You randomly choose a ball, hit it toward the flag, then ran

natima [27]

Answer:

1476

Step-by-step explanation:

Because

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

X = 35.5

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

$Z = \frac{35.5 - 26.751}{5.08}$

$Z = 1.72$

$Z = 1.72$ has a pvalue of 0.9573

X = 24.5

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

$Z = \frac{24.5 - 26.751}{5.08}$

$Z = -0.44$

$Z = -0.44$ has a pvalue of 0.3300

0.9573 - 0.3300 = 0.6273

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

(d) more than 40 will live beyond their 90th birthday

This is, using continuity correction, P(X > 40+0.5) = P(X > 40.5), which is 1 subtracted by the pvalue of Z when X = 40.5. So

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

$Z = \frac{40.5 - 26.751}{5.08}$

$Z = 2.71$

$Z = 2.71$ has a pvalue of 0.9966

1 - 0.9966 = 0.0034

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

6 0

3 years ago

Hese are the values in Paul’s data set.

andrew-mc [135]

Answer: See description below.

The residuals are the differences between the predicted values and the actual values. You will need to make a scatter plot of each difference.

Here are the five points that you will need to plot:
20 - 21 = -1 (1, -1)
17 - 16 = 1 (2, 1)
9 - 10 = -1 (4, -1)
6 - 5 = 1 (5, 1)
2 - 2 = 0 (6, 0)

8 0

3 years ago

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