Please answer this question now

How would the area of a triangle be affected if the base and height are dilated by 1/2

Sindrei [870]

Answer:

The area would reduce to 1/4 the original area

Step-by-step explanation:

The area of a triangle = (1/2)(base)(height) or bh/2

If the triangle base and height were to reduce by one-half, the formula would change to:

Area = (1/2)(1/2 base)(1/2 height) or bh/8. Which is a reduction by 1/4

6 0

3 years ago

X2 +<br><br> x + 36<br> Help me lol

Sergio039 [100]

Answer:

x²+12x+36 =0

Step-by-step explanation:

Hii, so I'm assuming your equation is equal to 0-->

x²+_x+36 =0

this could go many ways..

x²+12x+36 =0

which factor to (x+6)(x+6)

x²+15x+36 =0

which factor to (x+12)(x+3)

x²+13x+36 =0

which factor to (x+9)(x+4)

If I had to choose a answer.. I would go with the 1st option x²+12x+36 =0

Hope this helps, if there was more to this question, leave a comment and I can come back and help you ;)

8 0

3 years ago

Read 2 more answers

Simplify this algebraic expression.

Maru [420]

Answer:

D

Step-by-step explanation:

$y - \frac{3}{3} + 12$

$y - 1 + 12$

$y + 11$

4 0

3 years ago

How does the ratio of corresponding side lengths of the two quadrilaterals compare with the ratio of the areas of the quadrilate

Wittaler [7]

Answer:

The ratio of the areas would be the ratio of the side lengths squared.

Say the ratio of the side lengths is 1:2, the ratio of the areas would be 1:4. If the ratio of the sides were 2:3, the ratio of the areas would be 4:9.

Hope this helps

4 0

3 years ago

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