the triangle above is changed so that the Exterior angle measures 128 degrees. How does this affect angles B and C?

Solve y=x² +7 for x. A. x=y-7 B. x=+√y +7 C. x=y+7 D. x = t√y-7

Juli2301 [7.4K]

Answer:

D. x = t√y-7

Step-by-step explanation:

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4 0

2 years ago

Given the vectors u = <5,-1> and v = <1,12>, find -7u-3v. *

zepelin [54]

You multiply vectors and scalar simply by multiplying each component by that scalar:

$-7u = \langle-35,7\rangle,\quad -3v=\langle -3,-36\rangle$

Finally, you sum two vectors by summing the correspondent coordinates:

$-7u-3v = \langle-35,7\rangle+\langle -3,-36\rangle = \langle-38,-29\rangle$

4 0

2 years ago

a. Fill in the midpoint of each class in the column provided. b. Enter the midpoints in L1 and the frequencies in L2, and use 1-

Tresset [83]

Answer:

$\begin{array}{ccc}{Midpoint} & {Class} & {Frequency} & {64} & {63-65} & {1} & {67} & {66-68} & {11} & {70} & {69-71} & {8} &{73} & {72-74} & {7} & {76} & {75-77} & {3} & {79} & {78-80} & {1}\ \end{array}$

Using the frequency distribution, I found the mean height to be 70.2903 with a standard deviation of 3.5795

Step-by-step explanation:

Given

See attachment for class

Solving (a): Fill the midpoint of each class.

Midpoint (M) is calculated as:

$M = \frac{1}{2}(Lower + Upper)$

Where

$Lower \to$ Lower class interval

$Upper \to$ Upper class interval

So, we have:

Class 63-65:

$M = \frac{1}{2}(63 + 65) = 64$

Class 66 - 68:

$M = \frac{1}{2}(66 + 68) = 67$

When the computation is completed, the frequency distribution will be:

$\begin{array}{ccc}{Midpoint} & {Class} & {Frequency} & {64} & {63-65} & {1} & {67} & {66-68} & {11} & {70} & {69-71} & {8} &{73} & {72-74} & {7} & {76} & {75-77} & {3} & {79} & {78-80} & {1}\ \end{array}$

Solving (b): Mean and standard deviation using 1-VarStats

Using 1-VarStats, the solution is:

$\bar x = 70.2903$