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3 years ago

Divide 28 cans of soda into two groups so the ratio is 3 to 4

1 answer:

6 0

First, you add the two ratio numbers to figure out what to divide your 28 cans of soda by.

3=4=7
Next, you divide this amount into your total amount of cans of soda.

28/7=4.

This means that for each one in the ratio, there is 4 cans. So you multiply both numbers in the ratio by 4.

3*4 = 12
4*4=16

So, your final answer is 12:16, or 12 cans of soda in one group, and 16 cans of soda in the other.

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What is the slope-intercept form of the function described by this table?

Pavel [41]

Answer:

y = -4x +2

Step-by-step explanation:

As x-values increase by 1, y-values decrease by 4. The slope of the line is ...

... m = (change in y)/(change in x) = -4/1 = -4

We can use the first (x, y) pair as a point to use in the point-slope form of the equation of a line. That form can be written, for slope m and point (h, k) ...

... y = m(x -h) +k

using m = -4 and (h, k) = (1, -2), we can fill in the numbers to get ...

... y = -4(x -1) -2

... y = -4x +4 -2 . . . . eliminate parentheses

... y = -4x +2 . . . . . . slope-intercept form

_____

Alternate approach

After you recognize that a change in x of 1 gives a change in y of -4, you can work backward one step to find the table value for y corresponding to x=0. That will be -2+4 = +2. Now, you know both the slope (-4) and the y-intercept (+2), so you can write the equation directly from this knowledge:

... y = -4x +2

7 0

3 years ago

DOES ANYONE KNOW HOW TO DO THIS PROBLEM???? I AM NOT SURE HOW TO SOLVE THIS PROBLEM!!!!!!!!!!!!!!!!!

oksano4ka [1.4K]

option 2

option 3

option 1

Hope this helps my dude ^^

4 0

3 years ago

Read 2 more answers

Ten college students were randomly selected. Their grade point averages (GPAs) when they entered the program were between 3.5 a

solong [7]

Answer:

The 95% confidence interval for the true slope is (0.03985, 0.14206).

Step-by-step explanation:

For the regression equation:

$\hat y=\alpha +\hat \beta x$

The (1 - α)% confidence interval for the regression coefficient or slope $(\hat \beta )$ is:

$Ci=\hat \beta \pm t_{\alpha/2, (n-2)}\times SE(\hat \beta )$

The regression equation for current GPA (Y) of students based on their GPA's when entering the program (X) is:

$\hat Y=3.584756+0.090953 X$

The summary of the regression analysis is:

Predictor Coefficient SE t-stat p-value

Constant 3.584756 0.078183 45.85075 5.66 x 10⁻¹¹

Entering GPA 0.090953 0.022162 4.103932 0.003419

The regression coefficient and standard error are:

$\hat \beta = 0.090953\\SE (\hat \beta)=0.022162$

The critical value of t for 95% confidence level and 8 degrees of freedom is:

$t_{\alpha/2, n-2}=t_{0.05/2, 10-2}=t_{0.025, 8}=2.306$

Compute the 95% confidence interval for $(\hat \beta )$ as follows:

$CI=\hat \beta \pm t_{\alpha/2, (n-2)}\times SE(\hat \beta )\\=0.090953\pm 2.306\times 0.022162\\=0.090953\pm 0.051105572\\=(0.039847428, 0.142058572)\\\approx (0.03985, 0.14206)$