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Oduvanchick [21]
2 years ago
12

An experimenter records the results of a large sample survey on a histogram. Outcomes that lie far outside standard deviations o

f the data are
A: Medians
B: Statistically significant
C: Mean
Mathematics
1 answer:
Anna [14]2 years ago
6 0
Since a histogram is being used here to present the given data, this means that the information are presented in a bar form with different heights. Therefore, those outcomes that lie outside the standard deviations of the data would be considered as statistically significant. The answer to this is option B.
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2 years ago
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3 A buoy is 30 feet from the shore. You swim 3/5 the way to the buoy. How much farther do you have to swim to reach the buoy?​
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Answer:

12 feet

Step-by-step explanation:

30 divided by 5= 6

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6 times 2= how far you have left(12 feet)

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Rotate : energy :: stop:
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Craig has $1850 dollars in a bank account that he uses to make automatic payments of $400.73 on his car loan. If Craig stops mak
Ivenika [448]

Given:

Amount in the bank account = $1850

Monthly payment of can loan = $400.73

To find:

When would automatic payments make the value of the account zero?

Solution:

Craig stops making deposits to that account. So, amount $1850 in the bank account is used to make monthly payment of can loan.

On dividing the amount by monthly payment, we get

\dfrac{1850}{400.73}=4.61657

It means, the amount is sufficient for 4 payment but for the 5th payment the amount is not sufficient.

Therefore, the 5th automatic payments make the value of the account zero.

6 0
2 years ago
During the 2015-16 NBA season, J.J. Redick of the Los Angeles Clippers had a free throw shooting percentage of 0.901 . Assume th
ser-zykov [4K]

Answer: 0.5898

Step-by-step explanation:

Given :  J.J. Redick of the Los Angeles Clippers had a free throw shooting percentage of 0.901 .

We assume that,

The probability that .J. Redick makes any given free throw =0.901  (1)

Free throws are independent.

So it is a binomial distribution .

Using binomial probability formula, the probability of getting success in x trials :

P(X=x)^nC_xp^x(1-p)^{n-x}

, where n= total trials

p= probability of getting in each trial.

Let x be binomial variable that represents the number of a=makes.

n= 14

p= 0.901     (from (1))

The probability that he makes at least 13 of them will be :-

P(x\geq13)=P(x=13)+P(x=14)

=^{14}C_{13}(0.901)^{13}(1-0.901)^1+^{14}C_{14}(0.901)^{14}(1-0.901)^0\\\\=(14)(0.901)^{13}(0.099)+(1)(0.901)^{14}\ \ [\because\ ^nC_n=1\ \&\ ^nC_{n-1}=n ]\\\\\approx0.3574+0.2324=0.5898

∴ The required probability = 0.5898

5 0
2 years ago
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