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3 years ago

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A spinner with 6 equally sized slices has 2 red slices, 1 blue slice, and 3 yellow slices. The dial is spun and stops on a slice

at random. What is the probability that the dial stops on a red slice?

1 answer:

Dmitry_Shevchenko [17]3 years ago

5 0

Answer:

The probability is 2/6, or 1/3 (about 33%)

Step-by-step explanation:

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Two random samples are taken, with each group asked if they support a particular candidate. A summary of the sample sizes and pr

Minchanka [31]

Answer:

And we got $\alpha/2 =0.01$ so then the value for $\alpha=0.02$ and then the confidence level is given by: $Conf=1-0.02=0.98[/tex[ or 98%Step-by-step explanation:A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval". The margin of error is the range of values below and above the sample statistic in a confidence interval. Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean". [tex]p_1$ represent the real population proportion for 1

$\hat p_1 =0.768$ represent the estimated proportion for 1

$n_1=92$ is the sample size required for 1

$p_2$ represent the real population proportion for 2

$\hat p_2 =0.646$ represent the estimated proportion for 2

$n_2=95$ is the sample size required for 2

$z$ represent the critical value for the margin of error

The population proportion have the following distribution

$p \sim N(p,\sqrt{\frac{p(1-p)}{n}})$

The confidence interval for the difference of two proportions would be given by this formula

$(\hat p_1 -\hat p_2) \pm z_{\alpha/2} \sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1} +\frac{\hat p_2 (1-\hat p_2)}{n_2}}$

For this case we have the confidence interval given by: (-0.0313,0.2753). From this we can find the margin of erro on this way:

$ME= \frac{0.2753-(-0.0313)}{2}=0.1533$

And we know that the margin of erro is given by:

$ME=z_{\alpha/2} \sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1} +\frac{\hat p_2 (1-\hat p_2)}{n_2}}$

We have all the values except the value for $z_{\alpha/2}$

So we can find it like this:

$0.1533=z_{\alpha/2} \sqrt{\frac{0.768(1-0.768)}{92} +\frac{0.646 (1-0.646)}{95}}$

And solving for $z_{\alpha/2}$ we got:

$z_{\alpha/2}=2.326$

And we can find the value for $\alpha/2$ with the following excel code:

"=1-NORM.DIST(2.326,0,1,TRUE)"

And we got $\alpha/2 =0.01$ so then the value for $\alpha=0.02$ and then the confidence level is given by: $Conf=1-0.02=0.98$ or 98%

7 0

3 years ago

PLEASE ANSWER ASAP 45 POINTS!!!!!

sveta [45]

Im saying C is the awnser

8 0

3 years ago

Problem Solving

N76 [4]

you're absolutely correct.

each system is sold for $2150, that includes cost + markup, namely the markup is the surplus amount otherwise called "profit".

they sold 12 of those, 2150 * 12 = 25800

they had $4824.36 in profits from it, so if we subtract that from the sale price, we'll be left with the cost of all 12 systems

25800 - 4824.36 = 20975.64

that's the cost for all 12 systems sold, how many times does 12 go into 20975.64? 20975.64 ÷ 12 = 1747.97.

7 0

3 years ago

ryzh [129]

Answers:

☆☆☆☆☆☆☆☆aTriangle ZYX
bTriangle YZX
cTriangle XZY
dTriangle XYZ

4 0

3 years ago

There were 10 cards in a bag labeled 0-9 what is the probability of drawing a 3 and then a 5 if the first card is not replaced b

serg [7]

Answer:

1/9 chance

Step-by-step explanation:

6 0

3 years ago