Choose the unequal symbol that would make the math statement true

4 How many 3-cup servings are in 4 cups? A. 1/2 B. 2/ C. 4 D. 12

anygoal [31]

This questions stated differently so I’m assuming it’s 4?

7 0

1 year ago

What is the solution? pls help

loris [4]

Answer:

Solve by substitution

[y=5/2x-4]

[y= -x+3]

Substitute y = -x +3

[-x+3=5/2x-4]

Isolate x for -x+3=5/2x-4: x=2

For y = -x+3

Substitute x = 2

y= -2+3

Simplify

y=1

The solutions to the system of equations are:

y=1 x=2

3 0

2 years ago

My homework lesson 2 order of operations

djyliett [7]

Not sure what your question is but if you are learning the order of operations you can use the phrase "Please Excuse My Dear Aunt Sally"....

Parentheses
Exponents
Multiplication from left to right
Division from left to right
Addition from left to right
Subtraction from left to right

Hope this help :)

3 0

3 years ago

A jumper in the long-jump goes into the jump with a speed of 12 m/s at an angle of 20° above the horizontal. How far does the ju

weeeeeb [17]

Answer:

x = 9.5 m

Step-by-step explanation:

Given:

This problem is based on 2D motion kinematics equations:

$V_0 = 12 m/s$

$\theta = 20^0$

$g=9.8 m/s^2$

At maximum height:

$V_y = 0$

Components of initial velocity:

$V_0x = V_0 cos (\theta) =(12 )(cos (20^0)) = 11.3 m/s$

$V_0x = V_0 sin (\theta) = (12)(sin (20^0)) = 4.10 m/s$

To find time to reach maximum height:

$V_y = V_0y -gt$

Plugging in the known values:

0 = 4.10 - 9.8t

$t = \frac{-4.10}{-9.8}$

$t = 0.418 s$

Total time = 2t = 2(0.418) = 0.836 s

to find horizontal distance covered:

$x = (V_0x)(t)\\x = (11.3)(0.836)\\x = 9.5 m$

8 0

3 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