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2 years ago

5x-2y ≤ 20 can you guys solve this.

Mathematics

2 answers:

Levart [38]2 years ago

7 0

Your answer is y ≥ −10 + 5x/2

pychu [463]2 years ago

5 0

Answer:

Step-by-step explanation:

Let's solve for x.

5x−2y≤20

Step 1: Add 2y to both sides.

5x−2y+2y≤20+2y

5x≤2y+20

Step 2: Divide both sides by 5.

5x /5 ≤ 2y+20 /5

x≤ 2 /5 y+4

Answer:

x≤ 2 /5 y+4

Let's solve for y.

5x−2y≤20

Step 1: Add -5x to both sides.

5x−2y+−5x≤20+−5x

−2y≤−5x+20

Step 2: Divide both sides by -2.

−2y /−2 ≤ −5x+20

−2 y≥ 5 /2 x−10

Answer:

y≥ 5 /2 x−10

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Brian invests $10,000 in an account earning 4% interest, compounded annually for 10 years. Five years after Brian's initial inve

pentagon [3]

Answer:

Brian has $776 more account in his account than Chris.

Step-by-step explanation:

Compound interest Formula:

$A=P(1+r)^t$

$I$ = A-P

A= Amount after t years

P= Initial amount

r= Rate of interest

t= Time in year

Given that,

Brian invests $10,000 in an account earning 4% interest, compounded annually for 10 years.

Here P = $10,000 , r= 4%=0.04, t=10 years

The amount in his account after 10 years is

$A=10000(1+0.04)^{10}$

=$14802.44

≈$14802

Five years after Brian's investment,Chris invests $10,000 in an account earning 7% interest, compounded annually for 5 years.

Here P = $10,000 , r= 7%=0.07, t=5 years

The amount in his account after 5 years is

$A=10000(1+0.07)^{5}$

=$14025.51

≈$14026

From the it is cleared that Brian has $(14802-14026)=$776 more account in his account than Chris.

6 0

2 years ago

Suppose that a recent article stated that the mean time spent in jail by a first-time convicted burglar is 2.5 years. A study wa

4vir4ik [10]

Using the z-distribution, it is found that since the test statistic is greater than the critical value, it can be concluded that the mean length of jail time has increased.

At the null hypothesis, it is tested if the mean length of jail time is still of 2.5 years, that is:

$H_0: \mu = 2.5$

At the alternative hypothesis, it is tested if it has increased, that is:

$H_1: \mu > 2.5$

We have the standard deviation for the population, thus, the z-distribution is used. The test statistic is given by:

$z = \frac{\overline{x} - \mu}{\frac{\sigma}{\sqrt{n}}}$

The parameters are:

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

For this problem, the values of the parameters are: $\overline{x} = 3, \mu = 2.5, \sigma = 1.5, n = 26$

Hence, the value of the test statistic is:

$z = \frac{\overline{x} - \mu}{\frac{\sigma}{\sqrt{n}}}$

$z = \frac{3 - 2.5}{\frac{1.5}{\sqrt{26}}}$

$z = 1.7$

The critical value for a right-tailed test, as we are testing if the mean is greater than a value, with a significance level of 0.05, is of $z^{\ast} = 1.645$