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2 years ago

6

Two distinct integers, $x$ and $y$, are randomly chosen from the set $\{1,2,3,4,5,6,7,8,9,10\}$. What is the probability that $x

y-x-y$ is even?

1 answer:

Vaselesa [24]2 years ago

7 0

24 i really dont know what

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Two of the exterior angles of a triangle are $158^\circ$ and $99^\circ.$ Find the third exterior angle, in degrees.

Lisa [10]

Answer:

The third exterior angle is $103^o$

Step-by-step explanation:

Given

$\theta = 158^o$

$\alpha = 99^o$

Required

The third exterior angle $(\beta)$

The exterior angles of a triangle add up to 360 degrees.

So:

$\theta + \alpha + \beta = 360^o$

Make $\beta$ the subject

$\beta = 360^o - (\theta + \alpha)$

Substitute known values

$\beta = 360^o - (158^o + 99^o)$

$\beta = 103^o$

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2 years ago

The diameter of a circle is 16 cm. Find its area in terms of Pi

babymother [125]

Answer:

A= 201.06

Step-by-step explanation:

A=πr^2

d=2r

Input the information into those formulas. Im honestly not sure if thats the right answer but its better than nothing.

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2 years ago

Can someone help me pls with these?!

geniusboy [140]

Answer:

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2 years ago

The state education commission wants to estimate the fraction of tenth grade students that have reading skills at or below the e

marusya05 [52]

Answer:

The estimation for the proportion of tenth graders reading at or below the eighth grade level is given by:

$\hat p =\frac{955-812}{955}= 0.150$

$0.150 - 1.64 \sqrt{\frac{0.150(1-0.150)}{955}}=0.131$

$0.150 + 1.64 \sqrt{\frac{0.150(1-0.150)}{955}}=0.169$

And the 90% confidence interval would be given (0.131;0.169).

Step-by-step explanation:

We have the following info given:

$n= 955$ represent the sampel size slected

$x = 812$ number of students who read above the eighth grade level

The estimation for the proportion of tenth graders reading at or below the eighth grade level is given by:

$\hat p =\frac{955-812}{955}= 0.150$

The confidence interval for the proportion would be given by this formula

$\hat p \pm z_{\alpha/2} \sqrt{\frac{\hat p(1-\hat p)}{n}}$

For the 90% confidence interval the significance is $\alpha=1-0.9=0.1$ and $\alpha/2=0.05$ , with that value we can find the quantile required for the interval in the normal standard distribution and we got.

$z_{\alpha/2}=1.64$

And replacing into the confidence interval formula we got:

$0.150 - 1.64 \sqrt{\frac{0.150(1-0.150)}{955}}=0.131$

$0.150 + 1.64 \sqrt{\frac{0.150(1-0.150)}{955}}=0.169$

And the 90% confidence interval would be given (0.131;0.169).

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Answer:

Can you provide a picture?

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