A line has a slope of 2/3 and contains the point (-1,-5). Which is an equation of this line?

Sholpan [36]

The perimeter is the sum of all side lengths of a given shape, and can be computed by adding up all of the side lengths.

So in this case, it would be 15 x 2 + 13 x 2 the sum of them would be equal to the perimeter of the object, I believe.

3 0

3 years ago

Based on the graph above, the solution to the system if equations is (____,___)

sveta [45]

Answer:

1, -3

Step-by-step explanation:

4 0

3 years ago

If one of the 50 states is chosen at random, then a standard die is rolled, Find the probability of getting a state that starts

Black_prince [1.1K]

Answer:

The final answer is 2/15 or 0.133

Step-by-step explanation:

The US States that start with an "N" are:

Nebraska
Nevada
New Hampshire
New Jersey
New Mexico
New York
North Carolina
North Dakota

The probability of selecting a state that starts with an N is 8/50

The probability of rolling at least a two is 5/6.

If we multiply the two probabilities together, we get 2/15.

3 0

3 years ago

Fluoxetine, a generic anti-depressant, claims to have, on average, at least 20 milligrams of active ingredient. An independent l

san4es73 [151]

Answer:

$t=\frac{18.7-20}{\frac{5}{\sqrt{80}}}=-2.326$

$df=n-1=80-1=79$

$p_v =P(t_{(79)}$

Step-by-step explanation:

Information given

$\bar X=18.7$ represent the sample mean for the content of active ingredient

$s=5$ represent the sample standard deviation for the sample

$n=80$ sample size

$\mu_o =20$ represent the value that we want to test

t would represent the statistic

$p_v$ represent the p value for the test

System of hypothesis

We need to conduct a hypothesis in order to check if the true mean for the active agent is at least 20 mg, the system of hypothesis would be:

Null hypothesis: $\mu \geq 20$

Alternative hypothesis: $\mu < 20$

The statistic would be:

$t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}$ (1)

Now we can calculate the statistic:

$t=\frac{18.7-20}{\frac{5}{\sqrt{80}}}=-2.326$

P value

The degrees of freedom are calculated like this:

$df=n-1=80-1=79$

Since is a one left tailed test the p value would be:

$p_v =P(t_{(79)}$

6 0

3 years ago

A report on consumer financial literacy summarized data from a representative sample of 1,663 adult Americans. Based on data fro

Tom [10]

Answer:

a) 95% Confidence interval = (0.536, 0.584)

This means that we are 95% confident that between 53.6% and 58.4% of Americans would give themselves grade of A or B on their financial knowledge of personal finance.

b) The result of the 95% confidence interval agrees with the claim that majority of Americans would give themselves a grade of A or B on their financial knowledge of personal finance because the interval obtained for possible values that the population proportion of Americans that would give themselves a grade of A or B on their financial knowledge of personal finance lies completely on the side that is greater than a proportion of 0.50 which indicates that truly, majority of Americans would give themselves a grade of A or B on their financial knowledge of personal finance.

Step-by-step explanation:

Confidence Interval for the population proportion is basically an interval of range of values where the true population proportion can be found with a certain level of confidence.

Mathematically,

Confidence Interval = (Sample proportion) ± (Margin of error)

Sample proportion = proportion of Americans in the sample that would give themselves grade of A or B on their financial knowledge of personal finance = p = (934/1663) = 0.56

Margin of Error is the width of the confidence interval about the mean.

It is given mathematically as,

Margin of Error = (Critical value) × (standard Error)

Critical value at 95% confidence interval for sample size of 1663 is obtained from the z-tables as the sample size is large enough for the sample properties to approximate the population properties.

Critical value = 1.960 (from the z-tables)

Standard error of the mean = σₓ = √[p(1-p)/n]

p = 0.56

n = sample size = 1663

σₓ = √[0.56×0.44/1663] = 0.0121723457 = 0.01217

95% Confidence Interval = (Sample proportion) ± [(Critical value) × (standard Error)]

CI = 0.56 ± (1.96 × 0.01217)

CI = 0.56 ± 0.0238532

95% CI = (0.5361468, 0.5838532)

95% Confidence interval = (0.536, 0.584)

b) The result of the 95% confidence interval agrees with the claim that majority of Americans would give themselves a grade of A or B on their financial knowledge of personal finance because the interval obtained for possible values that the population proportion of Americans that would give themselves a grade of A or B on their financial knowledge of personal finance lies completely on the side that is greater than a proportion of 0.50 which indicates that truly, majority of Americans would give themselves a grade of A or B on their financial knowledge of personal finance.

Hope this Helps!!!

5 0

3 years ago