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Solnce55 [7]
4 years ago
9

Helppp what’s is itttttt

Mathematics
1 answer:
ra1l [238]4 years ago
6 0
-4 is the answer in this problem
You might be interested in
How much money would you need to deposit today at 7% annual interest
Monica [59]

Answer:

$33,090.

Step-by-step explanation:

The formula is A = P( 1 + r/2)^2n so we have:

50,000 = P(1 + 0.07/2)^(6*2)

50000 = P * 1.035^12

P = 50000/ 1.035^12

= $33,090 to the nearest dollar.

4 0
3 years ago
Derrick will need $39,500 in 10 years for college tuition. How much should his parents invest now at 9.5% annual interest, compo
erastova [34]

Answer:

His parents should invest $15,278.16 to reach this goal ⇒ 4th answer

Step-by-step explanation:

* Lets explain how to solve the problem

- Derrick will need $39,500 in 10 years for college tuition

∴ The future amount is $39,500

∴ The time for investment is 10 years

- P is the money his parents invest now at 9.5% annual interest,

 compounded daily

∴ The rate is 9.5% per year compounded daily

- The formula of the compounded interest is:

  A=P(1+\frac{r}{n})^{nt} , where

# A is the future value of money

# P is the value of investment

# r is the rate of interest in decimal

# t is the time of investment

# n is the period of the time

∵ A = $39,500

∵ t = 10

∵ r = 9.5/100 = 0.095

∵ n = 365 ⇒ compounded daily

- Lets use the formula above to find P

∴ 39500=P(1+\frac{0.095}{365})^{365*10}

∴ 39500=p(2.58539)

- Divide both sides by 2.58539

∴ P = $15278.16

∴ His parents should invest $15,278.16 to reach this goal

7 0
3 years ago
Please answer quickly its a test and i need to get a good grade on it, will give brainliest
Helga [31]

I think the answer is A because it I'm right it did double the size.

3 0
3 years ago
A random sample of 25 ACME employees showed the average number of vacation days taken during the year is 18.3 days with a standa
Norma-Jean [14]

Answer:

a) Null hypothesis:\mu \leq 15  

Alternative hypothesis:\mu > 15  

b) df=n-1=25-1=24  

For this case the p value is given p_v = 0.0392

If we compare the p value and the significance level given \alpha=0.05 we see that p_v so we can conclude that we have enough evidence to reject the null hypothesis, so we can conclude that the true mean is higher than 15 at 5% of signficance.  

c) Type I error, also known as a “false positive” is the error of rejecting a null  hypothesis when it is actually true. Can be interpreted as the error of no reject an  alternative hypothesis when the results can be  attributed not to the reality.

So for this case a type I of error would be reject the hypothesis that the true mean is less or equal than 15 and is actually true.

Step-by-step explanation:

Data given and notation  

\bar X=18.3 represent the sample mean

s=3.72 represent the sample standard deviation

n=25 sample size  

\mu_o =15 represent the value that we want to test

\alpha=0.05 represent the significance level for the hypothesis test.  

t would represent the statistic (variable of interest)  

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

Part a: State the null and alternative hypotheses.  

We need to conduct a hypothesis in order to check if the true mean for vacation days is higher than 15, the system of hypothesis would be:  

Null hypothesis:\mu \leq 15  

Alternative hypothesis:\mu > 15  

If we analyze the size for the sample is < 30 and we don't know the population deviation so is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:  

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

t-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".  

Part b: P-value  and conclusion

The first step is calculate the degrees of freedom, on this case:  

df=n-1=25-1=24  

For this case the p value is given p_v = 0.0392

Conclusion  

If we compare the p value and the significance level given \alpha=0.05 we see that p_v so we can conclude that we have enough evidence to reject the null hypothesis, so we can conclude that the true mean is higher than 15 at 5% of signficance.  

Part c

Type I error, also known as a “false positive” is the error of rejecting a null  hypothesis when it is actually true. Can be interpreted as the error of no reject an  alternative hypothesis when the results can be  attributed not to the reality.

So for this case a type I of error would be reject the hypothesis that the true mean is less or equal than 15 and is actually true.

3 0
3 years ago
For an angle θ with the point (−5, −12) on its terminating side, what is the value of cosine?
viva [34]

Answer:

option C. -\frac{5}{13}

Step-by-step explanation:

we have  that

The point (-5,-12) belong to the III quadrant

so

The value of the cosine is negative

Applying the Pythagoras Theorem

Find the value of the hypotenuse

h^{2}=5^{2}+12^{2}\\ h^{2} =25+144\\ h^{2}=169\\h=13\ units

The value of cosine of angle θ is the ratio between the side adjacent to angle θ and the hypotenuse

cos(\theta)=-\frac{5}{13}

8 0
3 years ago
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