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3 years ago

What is the least common multiple of 4,5,9, and 11

Mathematics

1 answer:

stepan [7]3 years ago

6 0

1,980 the anwser is right their at least that seems about right if I have calculated it correctly

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Use the properties of exponents to rewrite the expression.<br> (-3yz)(-3yz)(-3yz)(-3yz)

Lera25 [3.4K]

Answer:

Answer is 81y^4z^4

Step-by-step explanation:

The expression is (-3yz) (-3yz) (-3yz) (-3yz).

Since all the four terms have power 1:

(-3yz)^1 (-3yz)^1 (-3yz)^1 (-3yz)^1

We know that we can add the powers if the terms have same base

So,

=(-3yz)^1+1+1+1

=(-3yz)^4

=(-3)^4(y)^4(z)^4.

If the power is an even number than the negative sign changes into positive sign.

=3^4y^4z^4

=81y^4z^4

Thus the answer is 81y^4z^4....

3 0

3 years ago

There are 2 questions pls answer fast

Furkat [3]

Answer:

Q1: B. UHWXUQV

Q2: None of these answers equal 12 so I'll just tell you the answers for all of them in case you mistyped.

a. 66

b. 82

c. 120

d. 109

Im assuming you meant 120. If so, your answer is c.

5 0

3 years ago

the mean percentage of a population of people eating out at least once a week is 57% with a standard deviation of 3.5 %. assume

zhuklara [117]

Given:
Population proportion, $\mu _{p}$ = 57% = 0.57
Population standard deviation, σ = 3.5% = 0.035
Sample size, n = 40
Confidence level = 95%

The standard error is
$SE_{p} = \sqrt{ \frac{p(1-p)}{n} } = \sqrt{ \frac{0.57*0.43}{40} } =0.0783$

The confidence interval is
$\hat{p} \pm z^{*}SE_{p}$
where
$\hat{p}$ = sample proportion
z* = 1.96 at the 95% confidence lvvel

The sample proportion lies in the interval
(0.57-1.96*0.0783, 0.57+1.96*0.0783) = (0.4165, 0.7235)

Answer: Between 0.417 and 72.4), or between (41% and 72%)

8 0

3 years ago

A cheetah can run 70 miles per hour. What is this speed in feet per hour

GarryVolchara [31]

If I am correct the answer is 369,600 feet per hour.

7 0

3 years ago

Read 2 more answers

Consider a large shipment of 400 refrigerators, of which 40 have defective compressors. If X is the number among 15 randomly sel

Vlad [161]

Answer:

The required probability is 0.94

Step-by-step explanation:

Consider the provided information.

There are 400 refrigerators, of which 40 have defective compressors.

Therefore N = 400 and X = 40

The probability of defective compressors is:

$\frac{40}{400}=0.10$

It is given that If X is the number among 15 randomly selected refrigerators that have defective compressors,

That means n=15

Apply the probability density function.

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

We need to find P(X ≤ 3)

$P(X\leq3) =P(X=0)+P(X=1)+P(X=2)+P(X=3)\\P(X\leq3) =\frac{15!}{15!}(0.1)^0(1-0.1)^{15}+\frac{15!}{14!}(0.1)^1(1-0.1)^{14}+\frac{15!}{13!2!}(0.1)^2(1-0.1)^{13}+\frac{15!}{12!3!}(0.1)^3(1-0.1)^{12}\\$

$P(X\leq3) =0.944444369992\approx 0.94$

Hence, the required probability is 0.94

4 0

3 years ago