All categories

4 years ago

3(t-7) =6. Solve for the and show steps

Mathematics

1 answer:

Taya2010 [7]4 years ago

4 0

$3(t -7) = 6 \\ 3(t) + 3(-7)\\ 3(-7) = -21 \\ 3t = 3t \\ 3t - 21 = 6 \\Add21: 3t - 21 + 21 = 6 +21 \\ 3t = 21: Divide3 \\ \frac{3t}{3} cancel = \frac{27}{3}keep \\ t = 9$
________________________________________________________________
$8(h - 9) = 8 \\ 8h - 9 = 9 \\ 8h = 8h \\ 8(-9) = -72\\ 8h + (-72) = \\ 8h - 72 = 8 \\ Add72: 8h - 72 + 72 \\ = 8 + 72 \\ divide8 : \frac{8h}{8}cancel = \frac{80}{8} (keep) \\ h = 10$
________________________________________________________________
$9 + 2(z - 2) = 15 \\ 9+2(z)+ 2(-2)= 15 \\ 2(-2) = -4 \\ 9 +2z = ? \\ 9 +2z+4 = 15\\ Subtract5: 2z + 5 -5 \\ = 15 -5 \\ Divide2: 2z = 10 \\ \frac{2z}{2}cancel = \frac{10}{2} = 5 \\ z = 5$
:) :)
good luck on your assignment and enjoy your day

~ $MeIsKaitlyn:)$

You might be interested in

How would adding a score of zero affect the mean and mediam game scores

Nimfa-mama [501]

Answer:

Here is the complete question attached with.

The mean score would decrease more than the median score.

Step-by-step explanation:

The numbers for which we have to find the mean and median are:

$60,70,80,80,100$

Here the mean, $\frac{Sum\ of\ all\ observations}{Number\ of all\ observations} =\frac{60+70+80+80+100}{5} =78$

Median, $=80$ as median is the middle term if the observations are arranged in ascending order.

Now as the question says that we have to add a zero to see its effect.

So adding a zero we have

Mean $=\frac{0+60+70+80+80+100}{6} =65$

Median $= \frac{70+80}{2} =75$ ,as number of observations is even terms so we will add two middle numbers and divide it with $2$ .

So we can conclude that the mean is having more variation than the median.

Mean shows as variation of $78-65=13$ where as Median shows a variation of $80-75=5$ only.

So our final answer is option D that is "The mean score would decrease more than the median score."

8 0

3 years ago

M&M plain candies come in various colors. According to the M&M/Mars Department of Consumer Affairs, the distribution of

lyudmila [28]

Answer:

A) 0.12. Yes. Choosing a green and blue M&M is possible

B) 0.43. Yes. Choosing a yellow and red M&M is possible

C) 0.78

Step-by-step explanation:

First of all, the summation of the distribution of all colours is;

Σ(all colors ) = 22% + 20% + 23% + 10% + 6% + 6% + 13% = 100%, or 1.

Thus;

a) P(green candy or blue candy) is;

P(GREEN ∪ BLUE) = P(G) + P(BL)

P(GREEN ∪ BLUE) = 6%+6%

P(GREEN ∪ BLUE) = 12% or 0.12

Now, due to the fact that we have to choose ONE candy and only ONE candy at random, then they are mutually exclusive: Yes. Choosing a green and blue M&M is possible

b)P(yellow candy or red candy is;

P(YELLOW ∪ RED) = P(Y) + P(R)

P(YELLOW ∪ RED) = 20% + 23% = 43% or 0.43

Yes. Choosing a yellow and red M&M is possible

c) P(NOT PURPLE)

the probability of having a purple is;

P(PURPLE) = 22% or 0.22

So, the Probability of NOT having a PURPLE is 1 - 0.22 = 0.78

4 0

3 years ago

Find the distance<br> (-2, 1) and (4,3)

natima [27]