One of thr solutions to x2-2x-15=0 is x=-3 what is the other solution

Devon purchased one bag of dog food and three bags of cat food. Each bag of dog food weighs x pounds and each bag of cat food we

likoan [24]

Answer:

A pound of cat food will weigh 6.5 pounds

Step-by-step explanation:

Firstly, we need to get the value of x. Thus, let’s solve the equation at hand;

x + 3x/2 = 32.5

x + 1.5x = 32.5

2.5x = 32.5

x = 32.5/2.5

x = 13

Now from the question, we are made to know that a pound of cat food measures half of what a pound of dog food measures

Since we calculate a pound of dog food to weigh 13 pounds, a pound of cat food will weigh = 13/2 = 6.5 pounds

6 0

4 years ago

(0) 14 -7 0<br> | () 14 - 7 0

lyudmila [28]

Answer:

What is the question.

5 0

4 years ago

What is the solution to this system of equations? 4x + 5y = 7 3x – 2y = –12 The solution is .

babunello [35]

Answer:

(-2,3)

Step-by-step explanation:

8 0

3 years ago

Read 2 more answers

What is the first step in writing f(x) = 6x2 + 5 – 42x in vertex form?

atroni [7]

If we want to write the function in vertex form:
f ( x) = 6 x² + 5 - 42 x = 6 x² - 42 x + 5 = 6 ( x² - 7 x ) + 5 ...
First step is: C ) write the function in standard form ( y = ax² + b x + c )

8 0

4 years ago

Read 2 more answers

A data set lists earthquake depths. The summary statistics arenequals=500500,x overbarxequals=5.435.43km,sequals=4.864.86km. U

Eva8 [605]

Answer:

C. H 0:μ=5.00km H1:μ≠5.00km

$z=\frac{5.43-5.00}{\frac{4.86}{\sqrt{500}}}=1.978$

$p_v =2*P(z>1.978)=0.048$

If we compare the p value and the significance level given $\alpha=0.01$ we see that $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, so we can conclude that the true mean is nto significantly different from 5.00.

Step-by-step explanation:

Some previous concepts

The p-value is the probability of obtaining the observed results of a test, assuming that the null hypothesis is correct.

A z-test for one mean "is a hypothesis test that attempts to make a claim about the population mean(μ)".

The null hypothesis attempts "to show that no variation exists between variables or that a single variable is no different than its mean"

The alternative hypothesis "is the hypothesis used in hypothesis testing that is contrary to the null hypothesis"

Data given and notation

$\bar X=5.43$ represent the average for the sample

$s=4.86$ represent the sample standard deviation for the sample

$n=500$ sample size

$\mu_o =5.00$ represent the value that we want to test

$\alpha=0.01$ represent the significance level for the hypothesis test.

z would represent the statistic (variable of interest)

$p_v$ represent the p value for the test (variable of interest)

State the null and alternative hypotheses.

We need to conduct a hypothesis in order to check if the population mean is equal to 5.00, the system of hypothesis would be:

Null hypothesis: $\mu = 5.00$

Alternative hypothesis: $\mu \neq 5.00$

Since the sample size >30 and large, the estimator for the population deviation would be s, and we can use the z test to compare the actual mean to the reference value, and the statistic is given by:

$z=\frac{\bar X-\mu_o}{\frac{\sigma}{\sqrt{n}}}$ (1)

z-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".

Calculate the statistic

We can replace in formula (1) the info given like this:

$z=\frac{5.43-5.00}{\frac{4.86}{\sqrt{500}}}=1.978$

P-value

Since is a two tailed test the p value would be:

$p_v =2*P(z>1.978)=0.048$

In Excel we can us ethe following formula to find the p value "=2*(1-NORM.DIST(1.978;0;1;TRUE))"

Conclusion

If we compare the p value and the significance level given $\alpha=0.01$ we see that $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, so we can conclude that the true mean is nto significantly different from 5.00.

8 0

3 years ago