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2 years ago

The area of the top of a square table is 269 in2. What are the dimensions of the top?

Mathematics

1 answer:

dem82 [27]2 years ago

4 0

Answer:

$\sqrt{269}$ is ≈ 14.01

Step-by-step explanation:

If the table top is a square, to find the area, you would multiply one side times itself.

If you are given the area, and need to find the sides, you will take the square root of the area to find the side length.

Your question doesn't say to round, but 269 is not a perfect square.

$\sqrt{269}$ is ≈ 14.01

If the area of the base was 169, that is a perfect square. $\sqrt{169}$ = 13

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Solve this system of linear equations by substitution method

VikaD [51]

If I were you, I would select the first equation to put in y=mx+b form so that you can substitute it in to the second one. What you do is you subtract the x from the left side and bring it over to the right and then you divide everything by 5. The y=mx+b equation is y=-1/5x-7. Then, you substitute this equation into the bottom one so you are now working only with x's. You should distribute and combine like terms. When you distribute and clean it up, at the end you should get 18/5x-14=8. You then add 14 to get 22 on the right side and then multiply that by the reciprocal of the fraction. When you multiply 22 by 5/18, you should get a reduced fraction of 55/9. You then substitute this value into either equation to get the y-value. You should get a y-value of -260/45. Your answer as an ordered pair should be (55/9,-260/45).

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4 years ago

It is estimated that the population of the world is increasing at an average rate of 1.09%. The population was about 7,632,819,3

morpeh [17]

Answer:

The equation that represents the population after T years is

$P_{t} = 7,632,819,325 [1 +\frac{1.09}{100} ]^{T}$

Step-by-step explanation:

Population in the year 2018 ( P )= 7,632,819,325

Rate of increase R = 1.09 %

The population after T years is given by the formula

$P_{t} = P [1 +\frac{R}{100} ]^{T}$ -------- (1)

Where P = population in 2018

R = rate of increase

T = time period

Put the values of P & R in above equation we get

$P_{t} = 7,632,819,325 [1 +\frac{1.09}{100} ]^{T}$

This is the equation that represents the population after T years.

6 0

3 years ago

Can someone help me with both # 3

Sidana [21]

To answer this question I need a problem. Have a fabulous day,you are awesome!

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3 years ago

Five individuals from an animal population thought to be near extinction in a certain region have been caught, tagged, and relea

Talja [164]

Answer:

a) For this case the random variable X follows a hypergometric distribution.

b) $E(X)= n\frac{M}{N}=10 \frac{5}{25}=2$

$Var(X)=n \frac{M}{N}\frac{N-M}{N}\frac{N-n}{N-1}=10\frac{5}{25}\frac{25-5}{25}\frac{25-10}{25-1}=1$

c) $P(X=0)= \frac{(5C0)(25-5 C 10-0)}{25C10}=\frac{1*184756}{3268760}=0.0565$

d) $P(X=5)= \frac{(5C5)(25-5 C 10-5)}{25C10}=\frac{1*15504}{3268760}=0.00474$

Step-by-step explanation:

The hypergometric distribution is a discrete probability distribution that its useful when we have more than two distinguishable groups in a sample and the probability mass function is given by:

$P(X=k)= \frac{(MCk)(N-M C n-k)}{NCn}$

Where N is the population size, M is the number of success states in the population, n is the number of draws, k is the number of observed successes

The expected value and variance for this distribution are given by:

$E(X)= n\frac{M}{N}$

$Var(X)=n \frac{M}{N}\frac{N-M}{N}\frac{N-n}{N-1}$

a. What is the distribution of X?

For this case the random variable X follows a hypergometric distribution.

b. Compute the values for E(X) and Var(X)

For this case n=10, M=5, N=25, so then we can replace into the formulas like this:

$E(X)= n\frac{M}{N}=10 \frac{5}{25}=2$

$Var(X)=n \frac{M}{N}\frac{N-M}{N}\frac{N-n}{N-1}=10\frac{5}{25}\frac{25-5}{25}\frac{25-10}{25-1}=1$

c. What is the probability that none of the animals in the second sample are tagged?

So for this case we want this probability:

$P(X=0)= \frac{(5C0)(25-5 C 10-0)}{25C10}=\frac{1*184756}{3268760}=0.0565$

d. What is the probability that all of the animals in the second sample are tagged?

So for this case we want this probability:

$P(X=5)= \frac{(5C5)(25-5 C 10-5)}{25C10}=\frac{1*15504}{3268760}=0.00474$

4 0

3 years ago

Help me with this. Mathematics

s2008m [1.1K]

Answer:

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3 years ago