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2 years ago

Least common multiple of 72and40 step by step

Mathematics

1 answer:

Anton [14]2 years ago

8 0

Answer:4

Step-by-step explanation:first try different numbers and multiply them till you get your answer 4 times 18 is 72 and 4 time 10 is 40

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Rewrite the expression with a rational exponent as a radical expression.

Vilka [71]

Answer:

$\sqrt{5}$

Step-by-step explanation:

we know that

The "power rule" tells us that to raise a power to a power, just multiply the exponents

$(a^{m})^{n}=a^{m*n}$

we have

$(5^{\frac{3}{4}})^{\frac{2}{3}}$

Applying the "power rule"

$(5^{\frac{3}{4}})^{\frac{2}{3}}=5^{\frac{3}{4}*\frac{2}{3}}=5^{\frac{1}{2}}=\sqrt{5}$

5 0

3 years ago

You survey your class and record each students first name, last name, and age. You then make ordered pairs of (first name, age).

ExtremeBDS [4]

Answer: Yes

Step-by-step explanation:

It is a relation because each person will most likely have a different name and you can have plenty of the same outputs just as long as the input doesn't repeat at all.

5 0

3 years ago

Joe has one book each for algebra, geometry, history, psychology, Spanish, English and Physics in his locker. How many different

Alisiya [41]

Answer:

There are 35 different sets of 3 books Joe could choose

Step-by-step explanation:

* Lets explain how to solve the problem

- Combination is a collection of the objects where the order doesn't

matter

- The formula for the number of possible combinations of r objects from

a set of n objects is nCr = n!/r!(n-r)!

- n! = n(n - 1)(n - 2)................. × 1

Lets solve the problem

- Joe has one book each for algebra, geometry, history, psychology,

Spanish, English and Physics in his locker

∴ He has seven books in the locker

- He wants to chose three of them

∵ The order is not important when he chose the books

∴ We will use the combination nCr to find how many different sets

of three books he can choose

- The total number of books is 7

∴ n = 7

∵ He chooses 3 of them

∴ r = 3

∵ 7C3 = 7!/3!(7 - 3)! = 7!/3!(4!)

∴ $7C3=\frac{(7)(6)(5)(4)(3)(2)(1)}{[(3)(2)(1)][(4)(3)(2)(1)]}=35$

∴ 7C3 = 35

* There are 35 different sets of 3 books Joe could choose

3 0

3 years ago

Solve for c. 2(c +1) = 10

coldgirl [10]

Answer:

c=4

Step-by-step explanation:

Step 1: Simplify both sides of the equation.

2(c+1)=10

(2)(c)+(2)(1)=10(Distribute)

2c+2=10

Step 2: Subtract 2 from both sides.

2c+2−2=10−2

2c=8

Step 3: Divide both sides by 2.

2 = 8

c=4

8 0

3 years ago

Read 2 more answers

When the sample size and the sample proportion remain the same, a 90 percent confidence interval for a population proportion p w

deff fn [24]

Answer:

A 90% confidence interval for p will be narrower than the 99% confidence interval.

Step-by-step explanation:

The formula to compute the (1 - α) % confidence interval for a population proportion is:

$CI=\hat p\pm z_{\alpha /2}\sqrt{\frac{\hat p(1-\hat p)}{n}}$

Here $\hat p$ is the sample proportion.

The margin of error of the confidence interval is:

$MOE= z_{\alpha /2}\sqrt{\frac{\hat p(1-\hat p)}{n}}$

The MOE is dependent on:

Confidence level
Standard deviation
Sample size

The MOE is directly related to the confidence level and standard deviation.

So if any of the two increases then the MOE also increases, thus widening the confidence interval.

And the MOE is inversely related to the sample size.

So if the sample increases the MOE decreases and vice versa.

It is provided that the sample size and the sample proportion are not altered.

The critical value of z for 90% confidence level is:

$z_{\alpha/2}= z_{0.10/2}=z_{0.05}=1.645$

And the critical value of z for 99% confidence level is:

$z_{\alpha/2}= z_{0.05/2}=z_{0.05}=1.96$

So as the confidence level increases the critical value increases.

Thus, a 90% confidence interval for p will be narrower than the 99% confidence interval.

6 0

2 years ago