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alisha [4.7K]
3 years ago
8

A bag contains 7 blue, 4 yellow, 4 green, and 5 red marbles. Find the probability of choosing a green marble without looking. Ma

rk only one oval. Dood​
Mathematics
1 answer:
Naddik [55]3 years ago
3 0

Answer:

1/4

Step-by-step explanation:

total balls=7+4+4+5=20

probability of green marble=?

green marbles are 5 so,

       Probability=number of favourable outcomes / number of all possible outcomes

   total outcomes=20

green ball chances=5

       =5/20

          =1/4

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It's possible to build a triangle with a side lengths of 5, 5, and 10
yawa3891 [41]
Yeah, base is ten and the two sides are 5...
7 0
3 years ago
Read 2 more answers
The table showing the stock price changes for a sample of 12 companies on a day is contained in the Excel file below.
AfilCa [17]

Answer:

(a) The sample variance for the daily price change is 0.2501.

(b) The sample standard deviation for the daily price change is 0.5001.

(c) The 95% confidence interval estimates of the population variance is (0.1255, 0.7210).

Step-by-step explanation:

Let the random variable <em>X</em>  denote the stock price changes for a sample of 12 companies on a day.

The data provided is:

<em>X</em> = {0.82 , 1.44 , -0.07 , 0.41 , 0.21 , 1.33 , 0.97 , 0.30 , 0.14 , 0.12 , 0.42 , 0.15}

(a)

The formula to compute the sample variance for the daily price change is:

s^{2}=\frac{1}{n-1}\sum\limits^{12}_{i=1}{(X_{i}-\bar X)^{2}}

The sample mean is computed using the formula:

\bar X=\frac{1}{n}\sum\limits^{12}_{i=1}{X_{i}}

Consider the Excel output attached below.

In Excel the formula to compute the sample mean and sample variance are:

\bar X =AVERAGE(A2:A13)

s^{2} =VAR.S(A2:A13)

Thus, the sample variance for the daily price change is 0.2501.

(b)

The formula to compute the sample standard deviation for the daily price change is:

s=\sqrt{\frac{1}{n-1}\sum\limits^{12}_{i=1}{(X_{i}-\bar X)^{2}}}

Consider the Excel output attached below.

In Excel the formula to compute the sample standard deviation is:

s =STDEV.S(A2:A13)

Thus, the sample standard deviation for the daily price change is 0.5001.

(c)

The (1 - <em>α</em>)% confidence interval for population variance is:

CI=[\frac{(n-1)s^{2}}{\chi^{2}_{\alpha/2} } \leq \sigma^{2}\leq \frac{(n-1)s^{2}}{\chi^{2}_{1-\alpha/2} } ]

Compute the critical value of Chi-square for <em>α</em> = 0.05 and (n - 1) = (12 - 1) = 11 degrees of freedom as follows:

\chi^{2}_{\alpha/2, (n-1)}=\chi^{2}_{0.05/2,11}=21.920

\chi^{2}_{1-\alpha/2, (n-1)}=\chi^{2}_{(1-0.05/2),11}=\chi^{2}_{0.975,11}=3.816

*Use a Chi-square table.

Compute the 95% confidence interval estimates of the population variance as follows:

CI=[\frac{(n-1)s^{2}}{\chi^{2}_{\alpha/2} } \leq \sigma^{2}\leq \frac{(n-1)s^{2}}{\chi^{2}_{1-\alpha/2} } ]

     =[\frac{(12-1)\times 0.2501}{21.920 } \leq \sigma^{2}\leq \frac{(12-1)\times 0.2501}{3.816} ]

     =[0.125506\leq \sigma^{2}\leq 0.720938]\\\approx [0.1255, 0.7210]

Thus, the 95% confidence interval estimates of the population variance is (0.1255, 0.7210).

7 0
4 years ago
10 points. [ATTACHED]
SVETLANKA909090 [29]

Answer:

\overline{AD}\:=\:19.2

Step-by-step explanation:

2x+5=3x-2\quad \:\rightarrow \quad \:x=7\\\\\overline{BD}\:=\:3\left(7\right)-2\:=\:19\\\\Length\:of\:\overline{AD}\:=\sqrt{19^2+2.5^2}=19.2

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valentina_108 [34]
I think answer is 10
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Neko [114]

Answer:

<h3>\pi r^2h \\\pi (4^2) (12) \\192\pi</h3>

Step-by-step explanation:

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