64 is the correct answer. The numbers are being multiplied. 1×1=1, 2×2=4, 3×3=9, 4×4=16, etc.
Answer:
1. Slope: -3/4
2. Point-slope: y+4=-3/4(x+4)
3. Slope-intercept: y=-3/4x-7
4. Standard form: 3x+4y=-28
Step-by-step explanation:
1. To find the slope, use the slope formula, which is: y2-y1/x2-x1.
Plug the y-coordinates into the top part of the equation and the x-coordinates into the bottom part.
2--4/-4-4
2+4/-8 = 6/-8, or 3/-4
2. Use the slope and one of the x and y coordinates to put the equation into point-slope form. Recall that point-slope form is: y-y1=m(x-x1). Let's use the coordinates (4,-4).
y+4=-3/4(x+4)
3. To put the point-slope equation into slope-intercept form (y=mx+b), you need to distribute -3/4 to x+4 and subtract 4 from both sides.
When you distribute, the equation becomes:
y+4=-3/4x-3
Finally, when you subtract 4, the equation becomes:
y=-3/4x-7
4. Standard form is written as x+y= #.
To convert y=mx+b to standard form, first subtract -3/4x from both sides.
-3/4x + y = -7
Multiply everything by 4
3x + 4y = -28
Answer:
Step-by-step explanation:
From the given information:
Let's first compute the null and alternative hypothesis
Null hypothesis:

Alternative hypothesis:

The number of samples is half in a group of 60
i.e

the sample mean for sample 1
= 5.7
the standard deviation for sample 1
= 3.1
the sample mean for sample 2
= 3.9
the standard deviation for sample 2
= 2.7
degree of freedom for this test can be computed by using the formula:


df = 114.68
The test statistics can be computed as follows:


t = 1.07
Using the level of significance of 0.1, the P-value for the test statistics at the df of 114.68 is:
P-value = 0.143
Decision rule: To reject the null hypothesis if the level of significance is greater than the p-value.
Conclusion: We fail to reject the null hypothesis because the p-value is greater than the level of significance at 0.1.
Thus, it appears that the true average weight loss for the vegan diet does not exceeds that for the control diet by more than 1 kg.
50 like both out of 130 total
Divide: 50/130 = 5/13
The answer as a fraction is exactly 5/13
The approximate answer is 0.3846