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Anastaziya [24]

3 years ago

14

You pick a marble from this bag. What is the probability you will choose either a plain black or plain white marble? Simplify yo

ur answer.

1 answer:

exis [7]3 years ago

5 0

Answer:

50% chance for both

Step-by-step explanation:

If its two marbles then you have a 50% chance of picking up either the black or white. If four marbles are in the bag it would be 25%, if 10 then you have 10% and so on.

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Which equation has the same solution

Temka [501]

4x^2+48x=20
4x^2+48x+144=164
(2x+12).(2x+12)= + 59rt(164)
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8 0

3 years ago

What is x? <br><br> a) x=33.6 <br> b) x=21<br> c) x=24<br> d) x=26

kirill [66]

the answer is c) x=24

6 0

3 years ago

What is the solution to -x2 + 5x - 3

CaHeK987 [17]

The solution to the equation is $x=\frac{5-\sqrt{13}}{2}$ and $x=\frac{5+\sqrt{13}}{2}$

Explanation:

Given that the equation is $-x^2+5x-3=0$

We need to determine the solution of the equation.

The solution of the equation can be determined using the quadratic formula.

The quadratic formula is given by

$x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

Hence, from the equation, we have,

$a=-1,\:b=5,\:c=-3$

Substituting these values in the quadratic formula, we get,

$x=\frac{-5\pm \sqrt{5^2-4\left(-1\right)\left(-3\right)}}{2\left(-1\right)}$

Simplifying, we get,

$x=\frac{-5\pm \sqrt{25-12}}{-2}$

Simplifying the terms within the root, we get,

$x=\frac{-5\pm \sqrt{13}}{-2}$

Thus, the roots of the equation are $x=\frac{-5+ \sqrt{13}}{-2}$ and $x=\frac{-5- \sqrt{13}}{-2}$

Taking out the negative sign, we get,

$x=\frac{-(5- \sqrt{13})}{-2}$ and $x=\frac{-(5+ \sqrt{13})}{-2}$

Cancelling the negative sign, we get,

$x=\frac{5-\sqrt{13}}{2}$ and $x=\frac{5+\sqrt{13}}{2}$

Thus, the solutions of the equation are $x=\frac{5-\sqrt{13}}{2}$ and $x=\frac{5+\sqrt{13}}{2}$

7 0

4 years ago

A student at a four-year college claims that mean enrollment at four-year colleges is higher than at two-year colleges in the Un

WINSTONCH [101]

Answer:

Part 1: The statistic

$t=\frac{(\bar X_{1}-\bar X_{2})-\Delta}{\sqrt{\frac{\sigma^2_{1}}{n_{1}}+\frac{\sigma^2_{2}}{n_{2}}}}$ (1)

And the degrees of freedom are given by $df=n_1 +n_2 -2=35+35-2=68$

Replacing we got

$t=\frac{(5135-4436)-0}{\sqrt{\frac{783^2}{35}+\frac{553^2}{35}}}}=4.31$

Part 2: P value

Since is a right tailed test the p value would be:

$p_v =P(t_{68}>4.31)=0.000022 \approx 0.00002$

Comparing the p value we see that is lower compared to the significance level of 0.01 so then we can reject the null hypothesis and we can conclude that the mean for the four year college is significantly higher than the mean for the two year college and then the claim makes sense

Step-by-step explanation:

Data given

$\bar X_{1}=5135$ represent the mean for four year college

$\bar X_{2}=4436$ represent the mean for two year college

$s_{1}=783$ represent the sample standard deviation for four year college

$s_{2}=553$ represent the sample standard deviation two year college

$n_{1}=35$ sample size for the group four year college

$n_{2}=35$ sample size for the group two year college

$\alpha=0.01$ Significance level provided

t would represent the statistic (variable of interest)

System of hypothesis

We need to conduct a hypothesis in order to check if the mean enrollment at four-year colleges is higher than at two-year colleges in the United States , the system of hypothesis would be:

Null hypothesis: $\mu_{1}-\mu_{2}\leq 0$

Alternative hypothesis: $\mu_{1} - \mu_{2}> 0$

We can assume that the normal distribution is assumed since we have a large sample size for each case n>30. So then the sample mean can be assumed as normally distributed.

Part 1: The statistic

$t=\frac{(\bar X_{1}-\bar X_{2})-\Delta}{\sqrt{\frac{\sigma^2_{1}}{n_{1}}+\frac{\sigma^2_{2}}{n_{2}}}}$ (1)

And the degrees of freedom are given by $df=n_1 +n_2 -2=35+35-2=68$

Replacing we got

$t=\frac{(5135-4436)-0}{\sqrt{\frac{783^2}{35}+\frac{553^2}{35}}}}=4.31$

Part 2: P value

Since is a right tailed test the p value would be:

$p_v =P(t_{68}>4.31)=0.000022$

Comparing the p value we see that is lower compared to the significance level of 0.01 so then we can reject the null hypothesis and we can conclude that the mean for the four year college is significantly higher than the mean for the two year college and then the claim makes sense

6 0

3 years ago

Of the male tennis players surveyed more prefer

erica [24]

The answer would be A,B,A

6 0

3 years ago