30 is what percent of 64....

Need help due today will give brainiest and the first of a string of questions just like this one PLZ SHOW WORK NEEDED TO TURN I

sleet_krkn [62]

Answer:

Sorry bro you are going to get it

5 0

2 years ago

Solve each of the following for the indicated variable.a. 9 − 6 = 54 for yb. = 2 + 2 for l

marta [7]

Answer:

$\begin{gathered} y=\frac{9}{6}x-9 \\ l=\frac{P}{2}-w \end{gathered}$

Step-by-step explanation:

To solve the following equations, we need to isolate the variable asked using inverse operations to solve equations.

Remember that addition and subtraction are inverse operations, as multiplication and division.

Therefore,

$\begin{gathered} a\text{. Solve for y} \\ 6y=9x-54 \\ y=\frac{9}{6}x-\frac{54}{6} \\ y=\frac{9}{6}x-9 \end{gathered}$ $\begin{gathered} b\text{. Solve for l} \\ P=2l+2w \\ 2l=P-2w \\ l=\frac{P}{2}-\frac{2}{2}w \\ l=\frac{P}{2}-w \end{gathered}$

4 0

1 year ago

Please helpppppp

melomori [17]

Given Information:

Years = t = 35

Semi-annual deposits = P = $2,000

Compounding semi-annually = n = 2

Interest rate = i = 6.5%

Required Information

Accumulated amount = A = ?

Answer:

Accumulated amount = $515,827

Step-by-step explanation:

The future value of amount earned over period of 35 years and interest rate 6.5% with semi-annual deposits is given by

FV = PMT * ((1 + i/n)^nt - 1)/(i/n))

Where

n = 2

i = 0.065

t = 35

FV = 2000*((1 + 0.065/2)^2*35 - 1)/(0.065/2))

FV = 2,000*(257.91)

FV ≈ $515,827

Therefore, Anthony will have an amount of $515,827 when he retires in 35 years.

6 0

3 years ago

David wants to pack apples and bananas in his lunch. He has room for 16 fruit. He wants to pack three times as many apples as ba

S_A_V [24]

Answer:

David should pack 12 apples and 4 bananas

7 0

2 years ago

The following prices, in dollars, of 7.5-cubic-foot refrigerators were recorded from a random sample. 314 305 344 283 285 310 3

Mariana [72]

Answer:

$t=\frac{310.9-300}{\frac{31.09}{\sqrt{10}}}=1.109$

The degrees of freedom are given by:

$df=n-1=10-1=9$

And the p value would be given by:

$p_v =P(t_{9}>1.109)=0.148$

Since the p value is higher than the significance level of 0.05 we don't have enough evidence to conclude that the true mean is significantly higher than $300 because we FAIL to reject the null hypothesis.

Step-by-step explanation:

Information given

314 305 344 283 285 310 383 285 300 300

We can calculate the sample mean and deviation with the following formula:

$\bar X =\frac{\sum_{i=1}^n X_i}{n}$

$s =\sqrt{\frac{\sum_{i=1}^n (x_i -\bar X)^2}{n-1}}$

$\bar X=310.9$ represent the sample mean

$s=31.09$ represent the standard deviation for the sample

$n=10$ sample size

$\mu_o =300$ represent the value to test

$\alpha=0.05$ represent the significance level

t would represent the statistic

$p_v$ represent the p value

Hypothesis to test

We want to verify if the true mean is greater than 300, the system of hypothesis would be:

Null hypothesis: $\mu \leq 300$

Alternative hypothesis: $\mu > 300$

The statistic is given by:

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

Replacing the info given we got:

$t=\frac{310.9-300}{\frac{31.09}{\sqrt{10}}}=1.109$

The degrees of freedom are given by:

$df=n-1=10-1=9$

And the p value would be given by:

$p_v =P(t_{9}>1.109)=0.148$

Since the p value is higher than the significance level of 0.05 we don't have enough evidence to conclude that the true mean is significantly higher than $300 because we FAIL to reject the null hypothesis.

6 0

3 years ago