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3 years ago

42.6 is 66 2/3% of what number?

Mathematics

1 answer:

Elza [17]3 years ago

4 0

42.6 is 66 2/3% of 63.9

the answer is : 63.9

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The cost to mail a letter is a base charge of $0.42 plus $0.17 for each ounce. Write an equation that could be used to find how

tresset_1 [31]

Answer:

D) 0.42 + 0.17 x = 0.93

Step-by-step explanation:

0.93 - 0.42 = 0.51

0.51 / 0.17 = 3

3 ounces and the answer is D

3 0

2 years ago

Compare Look at these equations. Do you think they are all linear equations? Can they

Georgia [21]

Answer:

Yes they can all be written in y = mx + b. You just have to move the terms around.

Step-by-step explanation:

y = 2x -3, this is already in slope-intercept form

Now, y - 2 = x + 2: We can add 2 on both sides to cancel out the one on the left side:

y - 2 = x + 2

y - 2 + 2 = x + + 2

y = x + 4 <-- This is in y = mx + b form

Now the last one, 3x = 9 + 3y

We can first divide all terms by 3,

3x = 9 + 3y

/3 /3 /3

x = 3 + y: Then we can subtract 3 from both sides:

x - 3 = 3 + y - 3

x - 3 = y

These are all linear equations because none of the x's have bigger powers than 1. x^2 is a quadratic equation and x^3 is cubic equation.

6 0

2 years ago

Bryan B. kicks a 3kg soccer ball that accelerates at 9 m/s2. What is the force?

slega [8]

The answer is 27N

ur welcome

5 0

2 years ago

Assume that the paired data came from a population that is normally distributed. using a 0.05 significance level and dequalsxmin

Artemon [7]

"Assume that the paired data came from a population that is normally distributed. Using a 0.05 significance level and d = (x - y), find $\bar{d}$ , $s_{d}$ , the t-test statistic, and the critical values to test the claim that $\mu_{d} = 0$ "

You did not attach the data, therefore I can give you the general explanation on how to find the values required and an example of a random paired data.

For the example, please refer to the attached picture.

A) Find $\bar{d}$
You are asked to find the mean difference between the two variables, which is given by the formula:
$\bar{d} = \frac{\sum (x - y)}{n}$

These are the steps to follow:
1) compute for each pair the difference d = (x - y)
2) sum all the differences
3) divide the sum by the number of pairs (n)

In our example:
 $\bar{d} = \frac{6}{8} = 0.75$ 

B) Find $s_{d}$
You are asked to find the standard deviation, which is given by the formula:
 $s_{d} = \sqrt{ \frac{\sum(d - \bar{d}) }{n-1} }$

These are the steps to follow:
1) Subtract the mean difference from each pair's difference
2) square the differences found
3) sum the squares
4) divide by the degree of freedom DF = n - 1

In our example:
$s_{d} = \sqrt{ \frac{101.5}{8-1} }$
= √14.5
= 3.81

C) Find the t-test statistic.
You are asked to calculate the t-value for your statistics, which is given by the formula:
$t = \frac{(\bar{x} - \bar{y}) - \mu_{d} }{SE}$

where SE = standard error is given by the formula:
$SE = \frac{ s_{d} }{ \sqrt{n} }$

These are the steps to follow:
1) calculate the standard error (divide the standard deviation by the number of pairs)
2) calculate the mean value of x (sum all the values of x and then divide by the number of pairs)
3) calculate the mean value of y (sum all the values of y and then divide by the number of pairs)
4) subtract the mean y value from the mean x value
5) from this difference, subtract $\mu_{d}$
6) divide by the standard error

In our example:
SE = 3.81 / √8
= 1.346

The problem gives us $\mu_{d} = 0$ , therefore:
t = [(9.75 - 9) - 0] / 1.346
= 0.56

D) Find $t_{\alpha / 2}$
You are asked to find what is the t-value for a 0.05 significance level.

In order to do so, you need to look at a t-table distribution for DF = 7 and A = 0.05 (see second picture attached).

We find $t_{\alpha / 2} = 1.895$ 

Since our t-value is less than $t_{\alpha / 2}$ we can reject our null hypothesis!!

7 0

3 years ago

The height h (in feet) of an object dropped from a ledge after x seconds can be modeled by h(x)=−16x2+36 . The object is dropped

kakasveta [241]

Check the picture below.

$\bf ~~~~~~\textit{initial velocity in feet} \\\\ h(t) = -16t^2+v_ot+h_o \quad \begin{cases} v_o=\textit{initial velocity}&\\ \qquad \textit{of the object}\\ h_o=\textit{initial height}&\\ \qquad \textit{of the object}\\ h=\textit{object's height}&\\ \qquad \textit{at "t" seconds} \end{cases}$

so the object hits the ground when h(x) = 0, hmmm how long did it take to hit the ground the first time anyway?

$\bf h(x)=-16x^2+36\implies \stackrel{h(x)}{0}=-16x^2+36\implies 16x^2=36 \\\\\\ x^2=\cfrac{36}{16}\implies x^2 = \cfrac{9}{4}\implies x=\sqrt{\cfrac{9}{4}}\implies x=\cfrac{\sqrt{9}}{\sqrt{4}}\implies x = \cfrac{3}{2}~~\textit{seconds}$

now, we know the 2nd time around it hit the ground, h(x) = 0, but it took less time, it took 0.5 or 1/2 second less, well, the first time it took 3/2, if we subtract 1/2 from it, we get 3/2 - 1/2 = 2/2 = 1, so it took only 1 second this time then, meaning x = 1.

$\bf ~~~~~~\textit{initial velocity in feet} \\\\ h(x) = -16x^2+v_ox+h_o \quad \begin{cases} v_o=\textit{initial velocity}&0\\ \qquad \textit{of the object}\\ h_o=\textit{initial height}&\\ \qquad \textit{of the object}\\ h=\textit{object's height}&0\\ \qquad \textit{at "t" seconds}\\ x=\textit{seconds}&1 \end{cases} \\\\\\ 0=-16(1)^2+0x+h_o\implies 0=-16+h_o\implies 16=h_o \\\\[-0.35em] ~\dotfill\\\\ ~\hfill h(x) = -16x^2+16~\hfill$

quick info:

in case you're wondering what's that pesky -16x² doing there, is gravity's pull in ft/s².

4 0

2 years ago