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3 years ago

HELP! Please answer the question above tia

Mathematics

1 answer:

bonufazy [111]3 years ago

3 0

To get the Y values use the equation Y = MX and substitute the X and M values So for the first one you would get Y = -3 x -1 leaving Y as 3 then for the second you do Y = -3 x 0 giving you Y = 0 the third would be Y = -3 x 1 so Y would be -3 and you give the last one a shot >.O best of luck!

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PLEASE HELP IM STUCK. List the factors of the third term: 6n^3-n^2+5n/8-11

mafiozo [28]

Answer:

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Make sense?

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3 years ago

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3 years ago

What is the LCM of x^2+5 and x^2+10x+25?

avanturin [10]

Answer:

(x+5)²(x²+5)

Step-by-step explanation:

Given two functions x²+5 and x²+10x+25, to get their Lowest common factor, we need to to first factorize x²+10x+25

On factorising we have:

x²+5x+5x+25

= x(x+5) +5(x+5

= (x+5)(x+5)

= (x+5)²

The LCM can be calculated as thus

| x²+5, (x+5)²

x+5| x²+5, (x+5)

x+5| x²+5, 1

x²+5| 1, 1

The factors of both equation are x+5 × x+5 × x²+5

The LCM will be the product of the three functions i.e

(x+5)²(x²+5)

This hives the required expression.

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3 years ago

The Food and Drug Administration (FDA) produces a quarterly report called Total Diet Study. The FDA’s report covers more than 20

pav-90 [236]

Using the normal approximation to the binomial, it is found that:

The mean of X is of 52 samples.
The standard deviation of X is of 4.2661 samples.
0.0017 = 0.17% probability that less than half without any traces of pesticide.

--------------------------------

Binomial probability distribution

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

Can be approximated to a normal distribution, with mean given by:

$E(X) = np$

And standard deviation of:

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

Normal probability distribution

Z-score formula is used, which, in a set with mean $\mu$ and standard deviation $\sigma$ , the z-score of a measure X is given by:

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

Each z-score has an associated p-value, which measures the percentile of measure X.

--------------------------------

65% of samples had no pesticide, thus $p = 0.65$
80 samples, thus $n = 80$

The mean of X is of:

$E(X) = np = 0.65(80) = 52$

The standard deviation of X is of:

$\sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{80(0.65)(0.35)} = 4.2661$

Using the approximation and continuity correction, the probability is $P(X < 40 - 0.5) = P(X < 39.5)$ , which is the p-value of Z when X = 39.5. Thus: