Evaluate 13 - 0.75w + 8z when w = 12 and x = 1/2

Sam practices his guitar for 9 hours each week he practices each song for 3 1/2 hours. What is the greatest number of song he ca

liberstina [14]

Answer:

The greatest number of song he can practice in a week is 2.

Step-by-step explanation:

Suppose, the greatest number of song he can practice in a week is $x$

He practices each song for $3\frac{1}{2}$ hours. So, the total time required for practicing $x$ number of songs $= (3\frac{1}{2}*x)$ hours.

Given that, he practices his guitar for 9 hours each week.

So, the equation will be......

$3\frac{1}{2}*x= 9\\ \\ x= 9\div 3\frac{1}{2}\\ \\ x=9\div \frac{7}{2}=9 \times \frac{2}{7}=\frac{18}{7}= 2.571.... \approx 2$

(Rounded as the greatest whole number)

Thus, the greatest number of song he can practice in a week is 2.

4 0

2 years ago

Read 2 more answers

How to calculate the percentage?

miv72 [106K]

$\: \: \: \: \:$

How to calculate percentage of a number. Use the percentage formula: P% * X = Y

Convert the problem to an equation using the percentage formula: P% * X = Y.
P is 10%, X is 150, so the equation is 10% * 150 = Y.
Convert 10% to a decimal by removing the percent sign and dividing by 100: 10/100 = 0.10.

Multiply by 100 and add a percentage sign
0.876 = 0.876 * 100 = 87.6%

<h2>hope it helps</h2>

7 0

2 years ago

If Upper X overbar equals 62, Upper S equals 8, and n equals 36, and assuming that the population is normally distributed, c

marishachu [46]

Answer:

The 99% confidence interval would be given by (58.373;65.627)

We are 99% confident that the true mean for the variable of interest is between 58.373 and 65.627.

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

$\bar X=62$ represent the sample mean

$\mu$ population mean (variable of interest)

s=8 represent the sample standard deviation

n=36 represent the sample size

Part a: Confidence interval

The confidence interval for the mean is given by the following formula:

$\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}$ (1)

In order to calculate the critical value $t_{\alpha/2}$ we need to find first the degrees of freedom, given by:

$df=n-1=36-1=35$

Since the Confidence is 0.99 or 99%, the value of $\alpha=0.01$ and $\alpha/2 =0.005$ , and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.005,35)".And we see that $t_{\alpha/2}=2.72$

Now we have everything in order to replace into formula (1):

$62-2.72\frac{8}{\sqrt{36}}=58.373$

$62+2.72\frac{8}{\sqrt{36}}=65.627$

So on this case the 99% confidence interval would be given by (58.373;65.627)

We are 99% confident that the true mean for the variable of interest is between 58.373 and 65.627.

3 0

3 years ago

In figure above, if l1 | | l2 then value of x is: a) 40° b) 50° c) 80° d) 100°

cupoosta [38]

Answer:

its letter c so 80

Step-by-step explanation:

I hope this help

3 0

2 years ago

Suppose a spherical balloon grows in such a way that after t seconds, its volume is V = 4 sqrt(t) cm3. What is the volume of the

Arisa [49]

: You need to know the derivative of the sqrt function. Remember that sqrt(x) = x^(1/2), and that (d x^a)/(dx) = a x^(a-1). So (d sqrt(x))/(dx) = (d x^(1/2))/(dx) = (1/2) x^((1/2)-1) = (1/2) x^(-1/2) = 1/(2 x^(1/2)) = 1/(2 sqrt(x)).

There is a subtle shift in meaning in the use of t. If you say "after t seconds", t is a dimensionless quantity, such as 169. Also in the formula V = 4 sqrt(t) cm3, t is apparently dimensionless. But if you say "t = 169 seconds", t has dimension time, measured in the unit of seconds, and also expressing speed of change of V as (dV)/(dt) presupposes that t has dimension time. But you can't mix formulas in which t is dimensionless with formulas in which t is dimensioned.

Below I treat t as being dimensionless. So where t is supposed to stand for time I write "t seconds" instead of just "t".

Then (dV)/(d(t seconds)) = (d 4 sqrt(t))/(dt) cm3/s = 4 (d sqrt(t))/(dt) cm3/s = 4 / (2 sqrt(t)) cm3/s = 2 / (sqrt(t)) cm3/s.

Plugging in t = 169 gives 2/13 cm3/s.

4 0

3 years ago