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3 years ago

22 5. The probability of winning the game is 1/5 If you play the game

Mathematics

1 answer:

s2008m [1.1K]3 years ago

3 0

Answer:

twice. (2)

Step-by-step explanation:

1 in 5 times you play, you'll win. every time you play 5 times, you'll win one, so if you play 10 times (double) you'll win double (twice)

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2 years ago

Which expression is equivalent to b^-2/ab^-5?<br>a) a/b^5<br>b)1/ab^5<br>c)a^3b/1<br>d)b/a

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$\bf a^{-{ n}} \implies \cfrac{1}{a^{ n}}\qquad \qquad
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3 years ago

Read 2 more answers

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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3 years ago