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2 years ago

Which set is closed under subtraction?

Mathematics

2 answers:

ss7ja [257]2 years ago

8 0

The set of rational numbers.

Gemiola [76]2 years ago

7 0

the answer is the set of rational numbers

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Find an equation for the plane through the points p=(0,0,0), q=(0,1,0), and r=(1,2,3)

jek_recluse [69]

Hello,

Assume Ax+By+Cz+D=0 the plane 's equation.

p=(0,0,0)==>A*0+B*0+C*0+D=0==>D=0

q=(0,1,0)==>A*0+B*1+C*0+0=0==>B=0

r=(1,2,3)==>A*1+0*2+C*3+0=0==>A=-3C
Let C=1==>A=-3

An equation of the plane is -3x+z=0

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3 years ago

P(more than z=-2.59)is?

romanna [79]

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3 years ago

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3 years ago

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how far would a 4.0 cm tall object be located from a reflecting surface with a focal lenght that is equal to 5 cm if it is to cr

katrin [286]

Answer:

HERE IS YOUR ANSWER

Step-by-step explanation:

Use the mirror equation:

1/di + 1/do = 1/f

where di = -10 cm and f = +15 cm. (Note that di is negative if the image is virtual.)

Substitute and solve for do.

1/do + 1/(-10 cm) = 1/(15 cm)

1/do = 1/(15 cm) - 1/(-10 cm) = 5/(30 cm)

do = 6 cm

Hope it helps you

Regards,

Rachana

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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

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

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2 years ago