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3 years ago

What is the HIGHEST common factor of the set of numbers (54, 72, 90}?

Mathematics

2 answers:

suter [353]3 years ago

6 0

Answer:

Step-by-step explanation:

54/18=3

72/18=4

90/18=5

Travka [436]3 years ago

5 0

Answer:

Step-by-step explanation:

GCF of 54, 72, 90 is:

54÷3= 18

72÷4= 18

90÷5= 18

The GCF of 54, 72, and 90 is 18.

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Step-by-step explanation:

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3 years ago

What is the value of 5.4?

Galina-37 [17]

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3 years ago

g 78% of all Millennials drink Starbucks coffee at least once a week. Suppose a random sample of 50 Millennials will be selected

Tatiana [17]

Answer:

We know that n = 50 and p =0.78.

We need to check the conditions in order to use the normal approximation.

$np=50*0.78=39 \geq 10$

$n(1-p)=50*(1-0.78)=11 \geq 10$

Since both conditions are satisfied we can use the normal approximation and the distribution for the proportion is given by:

$p \sim N (p, \sqrt{\frac{p(1-p)}{n}})$

With the following parameters:

$\mu_ p = 0.78$

$\sigma_p = \sqrt{\frac{0.78*(1-0.78)}{50}}= 0.0586$

Step-by-step explanation:

Previous concepts

The binomial distribution is a "DISCRETE probability distribution that summarizes the probability that a value will take one of two independent values under a given set of parameters. The assumptions for the binomial distribution are that there is only one outcome for each trial, each trial has the same probability of success, and each trial is mutually exclusive, or independent of each other".

Solution to the problem

We know that n = 50 and p =0.78.

We need to check the conditions in order to use the normal approximation.

$np=50*0.78=39 \geq 10$

$n(1-p)=50*(1-0.78)=11 \geq 10$

Since both conditions are satisfied we can use the normal approximation and the distribution for the proportion is given by:

$p \sim N (p, \sqrt{\frac{p(1-p)}{n}})$

With the following parameters:

$\mu_ p = 0.78$

$\sigma_p = \sqrt{\frac{0.78*(1-0.78)}{50}}= 0.0586$

6 0

3 years ago

There are two calculus classes at your school classes have a class average of 75.5 the first class as a standard deviation of 10

DENIUS [597]

To compare the two classes, the Coefficient of Variation (COV) can be used.

The formula for COV is this:
C = s / x

where s is the standard deviation and x is the mean

For the first class:
C1 = 10.2 / 75.5
C1 = 0.1351 (13.51%)

For the second class:
C2 = 22.5 / 75.5
C2 = 0.2980 (29.80%)

The COV is a test of homogeneity. Looking at the values, the first class has more students having a grade closer to the average than the second class.

8 0

3 years ago

Which expression is eq to (125^2/125^4/3)

sweet [91]

Answer:

Step-by-step explanation:

So the denominator can be simplified to 5⁴ → $(5^3)^\frac{4}{3}$ and then the 3's cancel out and we are left with 5⁴

Now our new fraction is $\frac{125^{2} }{5^{4} } = \frac{15625}{625}$ which then simplifies to 25

5 0

3 years ago