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2 years ago

Solve. 0.25=−5+0.75+1.28

Mathematics

1 answer:

tatuchka [14]2 years ago

4 0

Answer:

the question is false

Step-by-step explanation:

0.25=-4.5+1.28 add -4.25 and 1.28 to get -2.97

0.25=-2.97 compare 0.25 and -2.9

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What gradient is shown in the graph

Lelu [443]

Answer:

-2/1 or -2

Step-by-step explanation:

in an equation this would be y=-2x-3

5 0

2 years ago

Convert the Cartesian equation (x 2 + y 2)2 = 4(x 2 - y 2) to a polar equation.

SashulF [63]

<u>ANSWER</u>

${r}^{2} = 4 \cos2\theta$

<u>EXPLANATION</u>

The Cartesian equation is

${( {x}^{2} + {y}^{2} )}^{2} = 4( {x}^{2} - {y}^{2} )$

We substitute

$x = r \cos( \theta)$

$y = r \sin( \theta)$

and

${x}^{2} + {y}^{2} = {r}^{2}$

This implies that

${( {r}^{2} )}^{2} = 4(( { r \cos\theta) }^{2} - {(r \sin\theta) }^{2} )$

Let us evaluate the exponents to get:

${r}^{4} = 4({ {r}^{2} \cos^{2}\theta } - {r}^{2} \sin^{2}\theta)$

Factor the RHS to get:

${r}^{4} = 4{r}^{2} ({ \cos^{2}\theta } - \sin^{2}\theta)$

Divide through by r²

${r}^{2} = 4 ({ \cos^{2}\theta } - \sin^{2}\theta)$

Apply the double angle identity

$\cos^{2}\theta -\sin^{2}\theta= \cos(2 \theta)$

The polar equation then becomes:

${r}^{2} = 4 \cos2\theta$

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

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