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sladkih [1.3K]
3 years ago
7

In your last 14 basketball games, you attempted 65 free throws and made 47. Find the experimental probability that you make a fr

ee throw. Write the probability as a percent, to the nearest tenth of a percent.
Mathematics
2 answers:
raketka [301]3 years ago
8 0

Answer:

Step-by-step explanation:

% = (sucesses / attempts) * 100

% = (47/65) * 100

% = 0.723 * 100

% = 72.3 %

Licemer1 [7]3 years ago
4 0

Answer: 72.3%  

Step-by-step explanation:

Given statements : The number of basketball games = 14

The number of free throws attempted = 65

The number of made = 47

Now, the experimental probability that you make a free throw is given by :-

\dfrac{\text{47}}{65}=0.72307\approx0.723

In percent , 0.723\times100=72.3\%

Hence, the  experimental probability that you make a free throw =72.3%  

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Hey mate!
You stuck?

8/34 would be your answer it would be the number of apple pies over the number of total pies.

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3 years ago
Glven: -7x < -21. Choose the solution set.​
SVEN [57.7K]

Answer:

x > 3

Step-by-step explanation:

-7x < -21

Divide each side by -7, remembering to flip the inequality

-7x/-7 > -21/-7

x > 3

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3 years ago
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3 years ago
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The lifetime of a certain type of battery is normally distributed with mean value 11 hours and standard deviation 1 hour. There
tankabanditka [31]

Answer:

The lifetime value needed is 11.8225 hours.

Step-by-step explanation:

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the zscore of a measure X is given by

Z = \frac{X - \mu}{\sigma}

After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we This p-value is the probability that the value of the measure is greater than X.

In this problem, we have that:

The lifetime of a certain type of battery is normally distributed with mean value 11 hours and standard deviation 1 hour. This means that \mu = 11, \sigma = 1.

What lifetime value is such that the total lifetime of all batteries in a package exceeds that value for only 5% of all packages?

This is the value of THE MEAN SAMPLE X when Z has a pvalue of 0.95. That is between Z = 1.64 and Z = 1.65. So we use Z = 1.645

Since we need the mean sample, we need to find the standard deviation of the sample, that is:

s = \frac{\sigma}{\sqrt{4}} = 0.5

So:

Z = \frac{X - \mu}{s}

1.645 = \frac{X - 11}{0.5}

X - 11 = 0.5*1.645

X = 11.8225

The lifetime value needed is 11.8225 hours.

3 0
3 years ago
a tank is 3/7 full of water.After removing 420 litres it became 12/35 full .how much can the tank hold when full? ​
insens350 [35]

Answer:

4900 Litres

Step-by-step explanation:

First we need a common demoninator:

3/7=15/35

Then we subtract the two to figure out how much 420 Litres is:

15/35-12/35=3/35

3/35=420 litres

Divide amount by the numerater for thow much 1/x is.

1/35=140

and multiply by the denomenator to get a full number on your fraction, and therefore a full tank.

140 x 35 = 4900

5 0
2 years ago
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