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2 years ago

One square has an area that is 10 cm2 larger than another. What is a reasonable domain for the area of the larger square?

Mathematics

2 answers:

vazorg [7]2 years ago

7 0

Answer:

$R>10$ is the required domain.

Step-by-step explanation:

We have been given two squares

Let the larger square area be x

We have been given the area of smaller square we need to find the domain of area of larger square.

Domian is the value that x can take in a function

Here, x is the area of the larger square

Since, area of smaller square is $10cm^2$

The area of larger circle has to be greater than $10cm^2$

The domain will be all real numbers grater than 10

Mathematically, $R>10$ is the required domain.

DENIUS [597]2 years ago

6 0

Given that the area can take any positive real number, the larger square hast to be more than 10 cm^2, then the domain for the area of the larger square is all the real numbers greater than 10 cm^2

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Problem 5 (4+4+4=12) We roll two fair 6-sided dice. Each one of the 36 possible outcomes is assumed to be equally likely. 1) Fin

tekilochka [14]

Answer:

$p(b) = \frac{1}{6}$

$p(k) = \frac{1}{3}$

$P(a) = \frac{1}{3}$

Step-by-step explanation:

Generally when two fair 6-sided dice is rolled the doubles are

(1 1) , ( 2 2) , (3 3) , (4 4) , ( 5 5 ), (6 6)

The total outcome of doubles is N = 6

The total outcome of the rolling the two fair 6-sided dice is

n = 36

Generally the probability that doubles (i.e., having an equal number on the two dice) were rolled is mathematically evaluated as

$p(b) = \frac{N}{n}$

$p(b) = \frac{6}{36}$

$p(b) = \frac{1}{6}$

Generally when two fair 6-sided dice is rolled the outcome whose sum is 4 or less is

(1, 1), (1, 2), (1, 3), (2, 1), (2, 2), (3, 1)

Looking at this outcome we see that there are two doubles present

The conditional probability that doubles were rolled is mathematically represented as

$p(k) = \frac{2}{6}$

$p(k) = \frac{1}{3}$

Generally when two fair 6-sided dice is rolled the number of outcomes that would land on different numbers is L = 30

And the number of outcomes that at least one die is a 1 is W = 10

The conditional probability that at least one die is a 1 is mathematically represented as

$P(a) = \frac{W}{L}$

=> $P(a) = \frac{10}{30}$

=> $P(a) = \frac{1}{3}$

3 0

1 year ago

The mean gross annual incomes of certified welders are normally distributed with the mean of $20,000 and a standard deviation of

grandymaker [24]

Answer:

At the 0.10 level of significance the z table gives critical values of -1.645 and 1.645 for two-tailed test.

Step-by-step explanation:

We are given that the mean gross annual incomes of certified welders are normally distributed with the mean of $20,000 and a standard deviation of $2,000.

The ship building association wishes to find out whether their welders earn more or less than $20,000 annually.

Let $\mu$ = mean gross annual incomes of certified welders

So, Null Hypothesis, $H_0$ : $\mu$ = $20,000

Alternate hypothesis, $H_A$ : $\mu \neq$ $20,000

Here, null hypothesis states that the mean income of welders is equal to $20,000.

On the other hand, alternate hypothesis states that the mean income of welders is not $20,000.

Also, the test statistics that would be used here is One-sample z test statistics as we know about the population standard deviation;

T.S. = $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ ~ N(0,1)

where, $\bar X$ = sample mean income

$\sigma$ = population standard deviation = $2,000

n = sample size

Now, at the 0.10 level of significance the z table gives critical values of -1.645 and 1.645 for two-tailed test.

3 0

2 years ago

It is claimed that 55% of marriages in the state of California end in divorce within the first 15 years. A large study was start

kykrilka [37]

Answer:

0.0045 = 0.45% probability that less than two of them ended in a divorce

Step-by-step explanation:

For each marriage, there are only two possible outcomes. Either it ended in divorce, or it did not. The probability of a marriage ending in divorce is independent of any other marriage. This means that the binomial probability distribution is used to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening.

55% of marriages in the state of California end in divorce within the first 15 years.

This means that $p = 0.55$

Suppose 10 marriages are randomly selected.

This means that $n = 10$

What is the probability that less than two of them ended in a divorce?

This is

$P(X < 2) = P(X = 0) + P(X = 1)$

In which

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 0) = C_{10,0}.(0.55)^{0}.(0.45)^{10} = 0.0003$

$P(X = 1) = C_{10,1}.(0.55)^{1}.(0.45)^{9} = 0.0042$

$P(X < 2) = P(X = 0) + P(X = 1) = 0.0003 + 0.0042 = 0.0045$

0.0045 = 0.45% probability that less than two of them ended in a divorce

8 0

1 year ago

Find the quotient. (5x4 – 3x2 4) ÷ (x 1)

Vlada [557]

Keywords:

Divide, polynomial, quotient, divisor, dividend, rest

For this case, we must find the quotient by dividing the polynomial $(5x ^ 4-3x ^ 2 + 4)\ by\ (x + 1)$ . We must build a quotient that, when multiplied by the divisor, eliminates the terms of the dividend until it reaches the rest, as shown in the attached figure. At the end of the division, to verify we must bear in mind that: