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3 years ago

Can someone plz help me this fast! Thank you

Mathematics

1 answer:

Mrrafil [7]3 years ago

4 0

Answer:

EF= 5 , DF= 6

Step-by-step explanation:

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I need help with this please :)

Marina CMI [18]

9514 1404 393

Answer:

all are Not Equivalent

Step-by-step explanation:

The coefficient of t³ in the target expression is -3. In order for that to be the result of simplifying any of the given expressions, they must have ( )^(t^3) in the denominator. None do, so none of the offered choices is equivalent to the given expression.

_____

The expressions simplify like this:

$\dfrac{(2^3)^{t^3}}{2^{5t}}=2^{3t^3-5t}\\\\\dfrac{8^{t^3}}{32^t}=2^{3t^3-5t}\\\\8^{t^3}\cdot32^t=2^{3t^3+5t}$

4 0

3 years ago

The table above represents what type of function?

exis [7]

Answer:

A?? because I think it's not linear because it's not proportional

7 0

3 years ago

Read 2 more answers

True or false 33 over 3 equals 11.3?

Leno4ka [110]

Answer:

fals it would be 11

Step-by-step explanation:

11 x 3 = 33

3 x 11 = 33

33 divide 11 = 3

33 divide 3 = 11

7 0

3 years ago

Read 2 more answers

Solve for x : x/-9=3

USPshnik [31]

Answer:

x=-27

Step-by-step explanation:

x/-9=3

x=3*-9=-27

7 0

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

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

5 0

3 years ago