Find the EXACT distance between the two points. (4,7) and (−4,9)

What is the value of n in the relationship 8n+9=-n+5

vitfil [10]

Subtract 5 from both sides.

$8n + 4 = -n$

Subtract 8n from both sides.

$4 = -9n$

Divide both sides by -9 to solve for n.

$- \frac{4}{9} = n$

The value of n is -4/9.

8 0

3 years ago

Suppose the lifetime of a cell phone battery is normally distributed with a mean of 36 months and a standard deviation of 2 mont

solong [7]

Answer:

They should guarantee the lifetime of their batteries for 32 months.

Step-by-step explanation:

Normal Probability Distribution:

Problems of normal distributions can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the z-score 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 p-value, we get the probability that the value of the measure is greater than X.

Mean of 36 months and a standard deviation of 2 months.

This means that $\mu = 36, \sigma = 2$

If the company wants to replace no more than 2% of all batteries, for how many months should they guarantee the lifetime of their batteries?

The guarantee should be the 2th percentile of lengths, which is X when Z has a pvalue of 0.02. So X when Z = -2.054.

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

$-2.054 = \frac{X - 36}{2}$

$X - 36 = -2.054*2$

$X = 31.89$

Rounding to the closest month, 32.

They should guarantee the lifetime of their batteries for 32 months.

3 0

2 years ago

I just need help plz

Licemer1 [7]

Answer:

1.4x 2 +7x−9

~#^××+$|₹-÷)=-₹2^_!&:'#^%&%&×(}/'?!24&0(86-

8 0

3 years ago

Read 2 more answers

Jeff is purchasing a car for $29,175 that will depreciate in value 7.5% each year.

beks73 [17]

Answer:

Step-by-step explanation:

figure that S*** out your self jeez

5 0

3 years ago

(x +y)^5 Complete the polynomial operation

Vesna [10]

Answer:

Please check the explanation!

Step-by-step explanation:

Given the polynomial

$\left(x+y\right)^5$

$\mathrm{Apply\:binomial\:theorem}:\quad \left(a+b\right)^n=\sum _{i=0}^n\binom{n}{i}a^{\left(n-i\right)}b^i$

$a=x,\:\:b=y$

$=\sum _{i=0}^5\binom{5}{i}x^{\left(5-i\right)}y^i$

so expanding summation

$=\frac{5!}{0!\left(5-0\right)!}x^5y^0+\frac{5!}{1!\left(5-1\right)!}x^4y^1+\frac{5!}{2!\left(5-2\right)!}x^3y^2+\frac{5!}{3!\left(5-3\right)!}x^2y^3+\frac{5!}{4!\left(5-4\right)!}x^1y^4+\frac{5!}{5!\left(5-5\right)!}x^0y^5$

solving

$\frac{5!}{0!\left(5-0\right)!}x^5y^0$

$=1\cdot \frac{5!}{0!\left(5-0\right)!}x^5$

$=1\cdot \:1\cdot \:x^5$

$=x^5$

also solving

$=\frac{5!}{1!\left(5-1\right)!}x^4y$

$=\frac{5}{1!}x^4y$

$=\frac{5x^4y}{1}$

$=5x^4y$

similarly, the result of the remaining terms can be solved such as

$\frac{5!}{2!\left(5-2\right)!}x^3y^2=10x^3y^2$

$\frac{5!}{3!\left(5-3\right)!}x^2y^3=10x^2y^3$

$\frac{5!}{4!\left(5-4\right)!}x^1y^4=5xy^4$

$\frac{5!}{5!\left(5-5\right)!}x^0y^5=y^5$

so substituting all the solved results in the expression

$=\frac{5!}{0!\left(5-0\right)!}x^5y^0+\frac{5!}{1!\left(5-1\right)!}x^4y^1+\frac{5!}{2!\left(5-2\right)!}x^3y^2+\frac{5!}{3!\left(5-3\right)!}x^2y^3+\frac{5!}{4!\left(5-4\right)!}x^1y^4+\frac{5!}{5!\left(5-5\right)!}x^0y^5$

$=x^5+5x^4y+10x^3y^2+10x^2y^3+5xy^4+y^5$

Therefore,

$\left(x\:+y\right)^5=x^5+5x^4y+10x^3y^2+10x^2y^3+5xy^4+y^5$

6 0

2 years ago