40% of what number is 34

Square root of 32-6√15

BlackZzzverrR [31]

Answer:

8 . 76209992276

Step-by-step explanation:

does this help

5 0

3 years ago

It is a timed test plz help i will give branlyiest

bixtya [17]

Answer:

5 edges and 4 vertices

Step-by-step explanation:

4 0

3 years ago

Multiply. (2x2 + 4x - 3)(x2 - 2x + 5)

BigorU [14]

For a moment, let $y=x^2-2x+5$ . Then by the distributive property,

$(2x^2+4x-3)p=2x^2p+4xp-3p$

Again by the distributive property,

$2x^2p=2x^2(x^2-2x+5)=2x^4-4x^3+10x^2$

$4xp=4x(x^2-2x+5)=4x^3-8x^2+20x$

$-3p=-3(x^2-2x+5)=-3x^2+6x-15$

Taking everything together, we get

$(2x^2+4x-3)(x^2-2x+5)=(2x^4-4x^3+10x^2)+(4x^3-8x^2+20x)+(-3x^2+6x-15)$
$(2x^2+4x-3)(x^2-2x+5)=2x^4-x^2+26x-15$

7 0

3 years ago

Read 2 more answers

BIG POINT FOR WHOEVER HELPS ME! PLEASE!!!!!Simplify and determine the coefficient of (-2/3X)(3Y).-X)

rewona [7]

You multiply like terms first, which means -2/3x*-x. It becomes 2/3x*3y. The coefficient of x is 2/3. The coefficient of y is 3. You can’t combine them because they are not like terms with the variables.

6 0

3 years ago

Suppose the horses in a large stable have a mean weight of 975lbs, and a standard deviation of 52lbs. What is the probability th

Lubov Fominskaja [6]

Answer:

0.8926 = 89.26% probability that the mean weight of the sample of horses would differ from the population mean by less than 15lbs if 31 horses are sampled at random from the stable

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 normally distributed random variable X, with mean $\mu$ and standard deviation $\sigma$ , the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$ .

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

In this problem, we have that:

$\mu = 975, \sigma = 52, n = 31, s = \frac{52}{\sqrt{31}} = 9.34$

What is the probability that the mean weight of the sample of horses would differ from the population mean by less than 15lbs if 31 horses are sampled at random from the stable?

pvalue of Z when X = 975 + 15 = 990 subtracted by the pvalue of Z when X = 975 - 15 = 960. So

X = 990

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

By the Central Limit Theorem

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

$Z = \frac{990 - 975}{9.34}$

$Z = 1.61$

$Z = 1.61$ has a pvalue of 0.9463

X = 960

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

$Z = \frac{960 - 975}{9.34}$

$Z = -1.61$

$Z = -1.61$ has a pvalue of 0.0537

0.9463 - 0.0537 = 0.8926

0.8926 = 89.26% probability that the mean weight of the sample of horses would differ from the population mean by less than 15lbs if 31 horses are sampled at random from the stable

7 0

3 years ago