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3 years ago

10

A basketball player makes 6 of 10 free throws in the first week of the season. If she shoots with the same accuracy for the next

week, how many of the 25 free throws she attempts will she make?

1 answer:

Alex777 [14]3 years ago

5 0

Answer:

15

Step-by-step explanation:

First divide: 6/10 ( to get a rate)

Next multiply: 25*0.6

Final answer: 15

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Suppose there are 4 defective batteries in a drawer with 10 batteries in it. A sample of 3 is taken at random without replacemen

SSSSS [86.1K]

Answer:

a.) 0.5

b.) 0.66

c.) 0.83

Step-by-step explanation:

As given,

Total Number of Batteries in the drawer = 10

Total Number of defective Batteries in the drawer = 4

⇒Total Number of non - defective Batteries in the drawer = 10 - 4 = 6

Now,

As, a sample of 3 is taken at random without replacement.

a.)

Getting exactly one defective battery means -

1 - from defective battery

2 - from non-defective battery

So,

Getting exactly 1 defective battery = ⁴C₁ × ⁶C₂ = $\frac{4!}{1! (4 - 1 )!}$ × $\frac{6!}{2! (6 - 2 )!}$

= $\frac{4!}{(3)!}$ × $\frac{6!}{2! (4)!}$

= $\frac{4.3!}{(3)!}$ × $\frac{6.5.4!}{2! (4)!}$

= $4$ × $\frac{6.5}{2.1! }$

= $4$ × $15$ = 60

Total Number of possibility = ¹⁰C₃ = $\frac{10!}{3! (10-3)!}$

= $\frac{10!}{3! (7)!}$

= $\frac{10.9.8.7!}{3! (7)!}$

= $\frac{10.9.8}{3.2.1!}$

= 120

So, probability = $\frac{60}{120} = \frac{1}{2} = 0.5$

b.)

at most one defective battery :

⇒either the defective battery is 1 or 0

If the defective battery is 1 , then 2 non defective

Possibility = ⁴C₁ × ⁶C₂ = 60

If the defective battery is 0 , then 3 non defective

Possibility = ⁴C₀ × ⁶C₃

= $\frac{4!}{0! (4 - 0)!}$ × $\frac{6!}{3! (6 - 3)!}$

= $\frac{4!}{(4)!}$ × $\frac{6!}{3! (3)!}$

= 1 × $\frac{6.5.4.3!}{3.2.1! (3)!}$

= 1× $\frac{6.5.4}{3.2.1! }$

= 1 × 20 = 20

getting at most 1 defective battery = 60 + 20 = 80

Probability = $\frac{80}{120} = \frac{8}{12} = 0.66$

c.)

at least one defective battery :

⇒either the defective battery is 1 or 2 or 3

If the defective battery is 1 , then 2 non defective

Possibility = ⁴C₁ × ⁶C₂ = 60

If the defective battery is 2 , then 1 non defective

Possibility = ⁴C₂ × ⁶C₁

= $\frac{4!}{2! (4 - 2)!}$ × $\frac{6!}{1! (6 - 1)!}$

= $\frac{4!}{2! (2)!}$ × $\frac{6!}{1! (5)!}$

= $\frac{4.3.2!}{2! (2)!}$ × $\frac{6.5!}{1! (5)!}$

= $\frac{4.3}{2.1!}$ × $\frac{6}{1}$

= 6 × 6 = 36

If the defective battery is 3 , then 0 non defective

Possibility = ⁴C₃ × ⁶C₀

= $\frac{4!}{3! (4 - 3)!}$ × $\frac{6!}{0! (6 - 0)!}$

= $\frac{4!}{3! (1)!}$ × $\frac{6!}{(6)!}$

= $\frac{4.3!}{3!}$ × 1

= 4×1 = 4

getting at most 1 defective battery = 60 + 36 + 4 = 100

Probability = $\frac{100}{120} = \frac{10}{12} = 0.83$

3 0

2 years ago

How many triangles can be constructed with the following measures?

sleet_krkn [62]

I got the answer by adding the triangles

4 0

2 years ago

In Applied Life Data Analysis (Wiley, 1982), Wayne Nelson presents the breakdown time of an insulating fluid between electrodes

ale4655 [162]

Answer:

The sample mean is $\bar{x}=14.371$ min.

The sample standard deviation is $\sigma = 18.889$ min.

Step-by-step explanation:

We have the following data set:

$\begin{array}{cccccccc}0.15&0.82&0.81&1.44&2.70&3.28&4.00&4.70\\4.96&6.49&7.25&8.03&8.40&12.15&31.89&32.47\\33.79&36.80&72.92&&&&&\end{array}$

The mean of a data set is commonly known as the average. You find the mean by taking the sum of all the data values and dividing that sum by the total number of data values.

The formula for the mean of a sample is

$\bar{x} = \frac{{\sum}x}{n}$

where, $n$ is the number of values in the data set.

$\bar{x}=\frac{0.15+0.82+0.81+1.44+2.7+3.28+4+4.7+4.96+6.49+7.25+8.03+8.4+12.15+31.89+32.47+33.79+36.80+72.92}{19}\\\\\bar{x}=14.371$

The standard deviation measures how close the set of data is to the mean value of the data set. If data set have high standard deviation than the values are spread out very much. If data set have small standard deviation the data points are very close to the mean.

To find standard deviation we use the following formula

$\sigma = \sqrt{ \frac{ \sum{\left(x_i - \overline{x}\right)^2 }}{n-1} }$

The mean of a sample is $\bar{x}=14.371$ .

Create the below table.

Find the sum of numbers in the last column to get.

$\sum{\left(x_i - \overline{X}\right)^2} = 6422.0982$

$\sigma = \sqrt{ \frac{ \sum{\left(x_i - \overline{x}\right)^2 }}{n-1} } = \sqrt{ \frac{ 6422.0982 }{ 19 - 1} } \approx 18.889$

7 0

3 years ago

Line / is a straight line.

sergejj [24]

Answer:

the first and last one

Step-by-step explanation:

6 0

2 years ago

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finlep [7]

Answer:

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Step-by-step explanation:

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6 0

2 years ago

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