Pleaseeee help me solve for X.

Suppose a principal amount of $2079 were invested into a simple interest account that pays an annual rate of 2.31% then the amou

Anna35 [415]

Answer:

The amount of interest earned at the end of 14 years would be $672.3486

Step-by-step explanation:

This is a simple interest problem.

The simple interest formula is given by:

$E = P*I*t$

In which E are the earnings, P is the principal(the initial amount of money), I is the interest rate(yearly, as a decimal) and t is the time.

In this problem, we have that:

$P = 2079, I = 0.0231, t = 14$

So

$E = 2079*0.0231*14 = 672.3486$

The amount of interest earned at the end of 14 years would be $672.3486

4 0

3 years ago

Help asap will mark brainliest

Roman55 [17]

Answer:

74

Step-by-step explanation:

Both angles x and 106 are supplementary which means

x+106=180

x=180-106

x=74

8 0

3 years ago

Read 2 more answers

Delta makes 12-volt car batteries. These batteries are known to be normally

blondinia [14]

Answer:

The probability that Delta car batteries last between three and four years

P(36≤X≤48) = 0.5188

The percentage of that Delta car batteries last between three and four years

P(3≤X≤4) = 52%

Step-by-step explanation:

Step(i):-

Given that the sample size n =12 -volt car batteries

Let 'X' be a Random variable in a normal distribution

Given that mean of the normal distribution = 45 months

Given that the Standard deviation of the normal distribution = 8months

Step(ii):-

Let X₁ = 3 years = 12 × 3 = 36 months

$Z_{1} = \frac{x_{1} -mean}{S.D} = \frac{36-45}{8} = -1.125$

Let X₂ = 4 years = 12 × 4 = 48 months

$Z_{2} = \frac{x_{2} -mean}{S.D} = \frac{48-45}{8} = 0.375$

Step(iii):-

The probability that Delta car batteries last between three and four years

P(36≤X≤48) = P(-1.125≤Z≤0.375)

= P(Z≤0.375) - P(Z≤-1.125)

= 0.5 +A(0.375) - (0.5-A(1.125)

= 0.5 + 0.1480 - (0.5 -0.3708)

= 0.1480 + 0.3708

= 0.5188

Final answer:-

The probability that Delta car batteries last between three and four years

P(36≤X≤48) = 0.5188

The percentage of that Delta car batteries last between three and four years

P(3≤X≤4) = 52%

5 0

3 years ago

A manufacturer of a new medication on the market for Alzheimer's disease makes a claim that the medication is effective in 65% o

NeX [460]

Answer:

$z=\frac{0.639 -0.65}{\sqrt{\frac{0.65(1-0.65)}{180}}}=-0.309$

$p_v =P(z$

So the p value obtained was a very high value and using the significance level given $\alpha=0.05$ we have $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can said that at 5% of significance the proportion of adults with the medication was effective is not significantly less than 0.65

Step-by-step explanation:

Data given and notation

n=180 represent the random sample taken

X=115 represent the adults with the medication was effective

$\hat p=\frac{115}{180}=0.639$ estimated proportion of adults with the medication was effective

$p_o=0.65$ is the value that we want to test

$\alpha=0.05$ represent the significance level

Confidence=95% or 0.95

z would represent the statistic (variable of interest)

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

Concepts and formulas to use

We need to conduct a hypothesis in order to test the claim that true proportion is less than 0.65.:

Null hypothesis: $p \geq 0.65$

Alternative hypothesis: $p < 0.65$

When we conduct a proportion test we need to use the z statisitc, and the is given by:

$z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}}$ (1)

The One-Sample Proportion Test is used to assess whether a population proportion $\hat p$ is significantly different from a hypothesized value $p_o$ .

Calculate the statistic

Since we have all the info requires we can replace in formula (1) like this:

$z=\frac{0.639 -0.65}{\sqrt{\frac{0.65(1-0.65)}{180}}}=-0.309$

Statistical decision

It's important to refresh the p value method or p value approach . "This method is about determining "likely" or "unlikely" by determining the probability assuming the null hypothesis were true of observing a more extreme test statistic in the direction of the alternative hypothesis than the one observed". Or in other words is just a method to have an statistical decision to fail to reject or reject the null hypothesis.

The significance level provided $\alpha=0.05$ . The next step would be calculate the p value for this test.

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

$p_v =P(z$

So the p value obtained was a very high value and using the significance level given $\alpha=0.05$ we have $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can said that at 5% of significance the proportion of adults with the medication was effective is not significantly less than 0.65

3 0

3 years ago

The cube of a positive number divided by the square of the same positive number

meriva

So, an example of this situation would be 3³÷3². When you divide like this, you simply would subtract 2 from 3 and that would be the new exponent. In this case, the exponent would be 1, so the expression is actually equal to whatever the positive number is. In another way, if you have 3×3×3 and you divide it by 3×3, you would have one three left, so that's your answer. Hope this helps!

3 0

3 years ago