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2 years ago

Can you find the difference, and put it in standard form?

Mathematics

2 answers:

Virty [35]2 years ago

4 0

Answer:

-5x^2 + 8 - 4x

Step-by-step explanation:

emove parentheses.

-2{x}^{2}+3-3{x}^{2}-4x+5

−2x

+3−3x

−4x+5

2 Collect like terms.

(-2{x}^{2}-3{x}^{2})+(3+5)-4x

(−2x

−3x

)+(3+5)−4x

3 Simplify.

-5{x}^{2}+8-4x

−5x

+8−4x

ExtremeBDS [4]2 years ago

3 0

Step-by-step explanation:

$\tt{( - 2 {x}^{2} + 3) - (3 {x}^{2} + 4x - 5)}$

Remember : While subtracting , the sign of each term of second expression changes & remove the parentheses.

⟶ $\tt{ - 2 {x}^{2} + 3 - 3 {x}^{2} - 4x + 5}$

Combine like terms. Like terms are those while have the same base.Only coefficients of like terms can be added or subtracted.

⟶ $\tt{ - 2 {x}^{2} - 3 {x}^{2} + 3 + 5 - 4x}$

⟶ $\tt{ - 5 {x}^{2} + 8 - 4x}$

In standard form , the expression should be written in a way that power of variables goes from highest to lowest.

$\red{ \boxed {\boxed{ \tt{Our \: final \: answer : \boxed{ \tt{ - 5 {x}^{2} - 4x + 8}}}}}}$

Hope I helped ! ♡

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A circle is centered at Q(1,-5) and has a radius of 5. Where does the point Y(4,-1) lie?

marta [7]

Answer:

Point Y (4, -1) lies on the circle.

Step-by-step explanation:

Distance = √[(x2-x1)² + (y2 - y1)²] = √[(4-1)² + (-1-(-5))²] = √[3² + 4²]= √25 = 5

Point Y (4, -1) lies on the circle, because the distance from the center to this point equals radius of the circle.

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3 years ago

The weight of an adult swan is normally distributed with a mean of 26 pounds and a standard deviation of 7.2 pounds. A farmer ra

Snezhnost [94]

Let $X$ denote the random variable for the weight of a swan. Then each swan in the sample of 36 selected by the farmer can be assigned a weight denoted by $X_1,\ldots,X_{36}$ , each independently and identically distributed with distribution $X_i\sim\mathcal N(26,7.2)$ .

You want to find

$\mathbb P(X_1+\cdots+X_{36}>1000)=\mathbb P\left(\displaystyle\sum_{i=1}^{36}X_i>1000\right)$

Note that the left side is 36 times the average of the weights of the swans in the sample, i.e. the probability above is equivalent to

$\mathbb P\left(36\displaystyle\sum_{i=1}^{36}\frac{X_i}{36}>1000\right)=\mathbb P\left(\overline X>\dfrac{1000}{36}\right)$

Recall that if $X\sim\mathcal N(\mu,\sigma)$ , then the sampling distribution $\overline X=\displaystyle\sum_{i=1}^n\frac{X_i}n\sim\mathcal N\left(\mu,\dfrac\sigma{\sqrt n}\right)$ with $n$ being the size of the sample.

Transforming to the standard normal distribution, you have

$Z=\dfrac{\overline X-\mu_{\overline X}}{\sigma_{\overline X}}=\sqrt n\dfrac{\overline X-\mu}{\sigma}$

so that in this case,

$Z=6\dfrac{\overline X-26}{7.2}$

and the probability is equivalent to

$\mathbb P\left(\overline X>\dfrac{1000}{36}\right)=\mathbb P\left(6\dfrac{\overline X-26}{7.2}>6\dfrac{\frac{1000}{36}-26}{7.2}\right)$
$=\mathbb P(Z>1.481)\approx0.0693$