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aivan3 [116]
3 years ago
9

A box at a miniature golf course contains contains 4 red golf balls, 8 green golf balls, and 7 yellow golf balls. What is the pr

obability of taking out a golf ball and having it be a red or a yellow golf ball? Express your answer as a percentage and round it to two decimal places.
Mathematics
1 answer:
Solnce55 [7]3 years ago
7 0

Answer:

 =57.89%

Step-by-step explanation:

The total number of golf ball is 4+8+7 = 19

P (red or yellow) = number of red or yellow

                              ------------------------------------

                                total number of golf balls

                           = 4+7

                              -----

                              19

                         =11/19

Changing this to a percent means changing it to a decimal and multiplying by 100%

                        = .578947368 * 100%

                         =57.8947368%

Rounding to two decimal places

                         =57.89%

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Answer: The correct answer is:  [D]:  "8" .

Step-by-step explanation:f y varies inversely with x;  then y = k/x ; with "k" being the "constant" in this equation.  

You can find the constant by substituting the values given: x = 16,  f(x) = y = 2 ;

So, f(x) = y = 2  = k/16 ;

→  2 = k / 16 ;

Solve for the constant; "k" ;

Multiply EACH SIDE of the equation by "16"

→  16* (2) = (k / 16) * 16 ;

→  32 = k ;

 ↔  k = 32 ;  

As such, we can write the equation:

y = k/ x ; as:

→  y = 32/ x ;

Since we are given:  "x = 4" ;  Plug in that value; and solve for "y" ;

y = 32/4 = 8 .

y = 8 .

y = f(x) ;

So;  f(x) = 8 ;  which is:  Answer choice:  [D]:  "8" .

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A research group needs to determine a 90% confidence interval for the mean repair cost for all car insurance small claims. From
lutik1710 [3]

Answer:

a) z = 1.645

b) The should sample at least 293 small claims.

Step-by-step explanation:

We have that to find our \alpha level, that is the subtraction of 1 by the confidence interval divided by 2. So:

\alpha = \frac{1-0.9}{2} = 0.05

Now, we have to find z in the Ztable as such z has a pvalue of 1-\alpha.

So it is z with a pvalue of 1-0.05 = 0.95, so z = 1.645, which means that the answer of question a is z = 1.645.

Now, find  the margin of error M as such

M = z*\frac{\sigma}{\sqrt{n}}

In which \sigma is the standard deviation of the population and n is the size of the sample.

(b) If the group wants their estimate to have a maximum error of $12, how many small claims should they sample?

They should sample at least n small claims, in which n is found when

M = 12, \sigma = 124.88. So

M = z*\frac{\sigma}{\sqrt{n}}

12 = 1.645*\frac{124.88}{\sqrt{n}}

12\sqrt{n} = 205.43

\sqrt{n} = \frac{205.43}{12}

\sqrt{n} = 17.12

\sqrt{n}^{2} = (17.12)^{2}

n = 293

The should sample at least 293 small claims.

8 0
2 years ago
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