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ser-zykov [4K]
2 years ago
6

GIVING BRAINLY!

Mathematics
1 answer:
stellarik [79]2 years ago
4 0
X-intercept: -9

Graph of the equation is shown below

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The perimeter of a rectangular concrete patio is 38 meters. The area is 90 square meters. What are the dimensions of the patio?
Nookie1986 [14]
I hope this helps you

6 0
3 years ago
**POINTS**Find the values of X and y
Andrej [43]

Answer:

y=105 for x

y=75

Step-by-step explanation:

2y-30=180                          180 (straight line)     180-105=75

+30      +30

2y=210

/2     /2

7 0
2 years ago
If Joe drives 186.83 miles on a business trip, and the reimbursement from his company is $13.08. At what rate is Joe's employer
kirill115 [55]

Answer:

7 cents/mile

Step-by-step explanation:

You are looking for a unit rate of cents per mile.

Change the dollar amount to cents, and divide by the number of miles.

$13.08 * (100 cents)/$ = 1308 cents

(1308 cents)/(183 miles) = 7.001 cents/mile

8 0
3 years ago
D^2(y)/(dx^2)-16*k*y=9.6e^(4x) + 30e^x
MA_775_DIABLO [31]
The solution depends on the value of k. To make things simple, assume k>0. The homogeneous part of the equation is

\dfrac{\mathrm d^2y}{\mathrm dx^2}-16ky=0

and has characteristic equation

r^2-16k=0\implies r=\pm4\sqrt k

which admits the characteristic solution y_c=C_1e^{-4\sqrt kx}+C_2e^{4\sqrt kx}.

For the solution to the nonhomogeneous equation, a reasonable guess for the particular solution might be y_p=ae^{4x}+be^x. Then

\dfrac{\mathrm d^2y_p}{\mathrm dx^2}=16ae^{4x}+be^x

So you have

16ae^{4x}+be^x-16k(ae^{4x}+be^x)=9.6e^{4x}+30e^x
(16a-16ka)e^{4x}+(b-16kb)e^x=9.6e^{4x}+30e^x

This means

16a(1-k)=9.6\implies a=\dfrac3{5(1-k)}
b(1-16k)=30\implies b=\dfrac{30}{1-16k}

and so the general solution would be

y=C_1e^{-4\sqrt kx}+C_2e^{4\sqrt kx}+\dfrac3{5(1-k)}e^{4x}+\dfrac{30}{1-16k}e^x
8 0
3 years ago
Which of the following Chi-Square tests can be used to determine if a data set follows a Normal distribution?
shepuryov [24]

Answer:

Answer = d. Chi-Square Goodness of Fit

Step-by-step explanation:

A decision maker may need to understand whether an actual sample distribution matches with a known theoretical probability distribution such as Normal distribution and so on. The Goodness-of-fit Test is a type of Chi-Square test that can be used to determine if a data set follows a Normal distribution and how well it fits the distribution. The Chi-Square test for Goodness-of-fit enables us to determine the extent to which theoretical probability distributions coincide with empirical sample distribution. To apply the test, a particular theoretical distribution is first hypothesized for a given population and then the test is carried out to determine whether or not the sample data could have come from the population of interest with hypothesized theoretical distribution. The observed frequencies or values come from the sample and the expected frequencies or values come from the theoretical hypothesized probability distribution. The Goodness-of-fit now focuses on the differences between the observed values and the expected values. Large differences between the two distributions throw doubt on the assumption that the hypothesized theoretical distribution is correct and small differences between the two distributions may be assumed to be resulting from sampling error.

6 0
3 years ago
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