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guajiro [1.7K]
3 years ago
15

All of the following are equivalent, except _____.

Mathematics
2 answers:
earnstyle [38]3 years ago
8 0
The correct answer would be b) (12 + 3x)x
8_murik_8 [283]3 years ago
5 0
All are equivalent except for b - because 12 * x = 12x but 3x * x becomes 3x^2. <span />
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I’m about to cry from frustration. Help please before I die
kvv77 [185]

Answer:

\frac{1}{3}x+16=37

63 Jelly Beans

Step-by-step explanation:

The unknown is the number of jelly beans originally in the bag or x

First he had x jelly beans

The he ate one-third of them

\frac{1}{3}x

He then ate 16 more jelly beans

\frac{1}{3}x+16

This was equal to 37 jelly beans

\frac{1}{3}x+16=37

This is the equation

Now solve for x

Subtract 16 from both sides

\frac{1}{3}x=21

Multiply both sides by 3

x=63

63 Jelly Beans

7 0
2 years ago
Read 2 more answers
Kevin is 3 times as old as Daniel. 4 years ago, Kevin was 5 times as old as Daniel.
rusak2 [61]

Answer:

24

Step-by-step explanation:

4 years ago, Daniel would've been 4 and Kevin would be 20, so Kevin would've been 5 times as old as Daniel. And 8 x 3 = 24. So, Kevin is 24.

4 0
3 years ago
Read 2 more answers
While conducting a test of modems being manufactured, it is found that 10 modems were faulty out of a random sample of 367 modem
Kitty [74]

Answer:

We conclude that this is an unusually high number of faulty modems.

Step-by-step explanation:

We are given that while conducting a test of modems being manufactured, it is found that 10 modems were faulty out of a random sample of 367 modems.

The probability of obtaining this many bad modems (or more), under the assumptions of typical manufacturing flaws would be 0.013.

Let p = <em><u>population proportion</u></em>.

So, Null Hypothesis, H_0 : p = 0.013      {means that this is an unusually 0.013 proportion of faulty modems}

Alternate Hypothesis, H_A : p > 0.013      {means that this is an unusually high number of faulty modems}

The test statistics that would be used here <u>One-sample z-test</u> for proportions;

                             T.S. =  \frac{\hat p-p}{\sqrt{\frac{p(1-p)}{n} } }  ~  N(0,1)

where, \hat p = sample proportion faulty modems= \frac{10}{367} = 0.027

           n = sample of modems = 367

So, <u><em>the test statistics</em></u>  =  \frac{0.027-0.013}{\sqrt{\frac{0.013(1-0.013)}{367} } }

                                     =  2.367

The value of z-test statistics is 2.367.

Since, we are not given with the level of significance so we assume it to be 5%. <u>Now at 5% level of significance, the z table gives a critical value of 1.645 for the right-tailed test.</u>

Since our test statistics is more than the critical value of z as 2.367 > 1.645, so we have sufficient evidence to reject our null hypothesis as it will fall in the rejection region due to which <u><em>we reject our null hypothesis</em></u>.

Therefore, we conclude that this is an unusually high number of faulty modems.

6 0
3 years ago
H E L P!!! Simplify the expression -4d+3-8+6+d-1
ser-zykov [4K]

-4d+3-8+6+d-1 =

-3d + 0 =

-3d

6 0
3 years ago
The school sold 200 tickets to play. Student tickets are $4 and adult tickets are $6. If they made a total $930 how many of each
r-ruslan [8.4K]

200 + \div 4 =
50
200 \div 6 =
33 \infty
There was 50 student tickets sold for the play.
There was 33 infinity adult tickets sold for the play

5 0
3 years ago
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