All categories

3 years ago

A rectangle has side lengths 2x + 3 and 5x - 2.

Mathematics

1 answer:

xz_007 [3.2K]3 years ago

5 0

Answer:

10x² + 11x - 6

Step-by-step explanation:

The area (A) is the product of the sides, that is

A = (2x + 3)(5x - 2) ← expand using FOIL

= 10x² - 4x + 15x - 6 ← collect like terms

= 10x² + 11x - 6

You might be interested in

Which situation could be represented by an exponential function select each correct answer

ANTONII [103]

Answer:

The first and third options are the examples of exponential functions.

Step-by-step explanation:

When a quantity is compounded after a certain interval of time at a certain rate, then we can assume that the situation can be represented by an exponential function.

In the first option: An event organizer finds each year's attendance for the past five years is about $\frac{4}{5}$ of previous year's attendance.

So, here the total attendance is compounding every year by a factor $\frac{4}{5}$ of previous year's attendance.

Again, in the third case: The total population is increasing by about 7.5% each year.

Hence, the population is compounded every year by 7.5% of the previous year's population.

Therefore, the first and third options are examples of exponential functions. (Answer)

3 0

2 years ago

An advertisement for a popular weight-loss clinic suggests that participants in its new diet program lose, on average, more than

Sedbober [7]

Testing the hypothesis, it is found that:

The null hypothesis is: $H_0: \mu \leq 10$

The alternative hypothesis is: $H_1: \mu > 10$

The critical value is: $t_c = 1.74$

The decision rule is:

If t < 1.74, we do not reject the null hypothesis.
If t > 1.74, we reject the null hypothesis.

Since t = 1.41 < 1.74, we do not reject the null hypothesis, that is, it cannot be concluded that the mean weight loss is of more than 10 pounds.

Item a:

At the null hypothesis, it is tested if the mean loss is of at most 10 pounds, that is:

$H_0: \mu \leq 10$

At the alternative hypothesis, it is tested if the mean loss is of more than 10 pounds, that is:

$H_1: \mu > 10$

Item b:

We are having a right-tailed test, as we are testing if the mean is more than a value, with a significance level of 0.05 and 18 - 1 = 17 df.

Hence, using a calculator for the t-distribution, the critical value is: $t_c = 1.74$ .

Hence, the decision rule is:

If t < 1.74, we do not reject the null hypothesis.
If t > 1.74, we reject the null hypothesis.

Item c:

We have the standard deviation for the sample, hence the t-distribution is used. The test statistic is given by:

$t = \frac{\overline{x} - \mu}{\frac{s}{\sqrt{n}}}$

The parameters are:

$\overline{x}$ is the sample mean.
$\mu$ is the value tested at the null hypothesis.
s is the standard deviation of the sample.
n is the sample size.

For this problem, we have that:

$\overline{x} = 10.8, \mu = 10, s = 2.4, n = 18$

Thus, the value of the test statistic is:

$t = \frac{\overline{x} - \mu}{\frac{s}{\sqrt{n}}}$

$t = \frac{10.8 - 10}{\frac{2.4}{\sqrt{18}}}$

$t = 1.41$

Since t = 1.41 < 1.74, we do not reject the null hypothesis, that is, it cannot be concluded that the mean weight loss is of more than 10 pounds.

A similar problem is given at brainly.com/question/25147864

3 0

2 years ago

What is a solution to x ≥ 20

Triss [41]

Answer:

Step-by-step explanation:

Any number equal to or greater than 20 is a solution. 22 would be a solution.

4 0

2 years ago

Help me please please now asap

Oduvanchick [21]

120 different 3digit numbers

6 0

3 years ago

The centers for disease control and prevention reported that 25% of baby boys 6-8 months old in the united states weigh more tha

melisa1 [442]

Answer:

0.1971 ( approx )

Step-by-step explanation:

Let X represents the event of weighing more than 20 pounds,

Since, the binomial distribution formula is,

$P(x)=^nC_r p^r q^{n-r}$

Where, $^nC_r=\frac{n!}{r!(n-r)!}$

Given,

The probability of weighing more than 20 pounds, p = 25% = 0.25,

⇒ The probability of not weighing more than 20 pounds, q = 1-p = 0.75

Total number of samples, n = 16,

Hence, the probability that fewer than 3 weigh more than 20 pounds,

$P(X$

$=^{16}C_0 (0.25)^0 (0.75)^{16-0}+^{16}C_1 (0.25)^1 (0.75)^{16-1}+^{16}C_2 (0.25)^2 (0.75)^{16-2}$

$=(0.75)^{16}+16(0.25)(0.75)^{15}+120(0.25)^2(0.75)^{14}$

$=0.1971110499$

$\approx 0.1971$

3 0

3 years ago