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lara [203]
3 years ago
11

After 10 weeks of work, bob has $1200. After 15 weeks of work Bob has $1800. After 20 weeks of work Bob has $2400. Write the fir

s four terms of a sequence that represents this situation. Writ and equation to model the situation
Mathematics
1 answer:
lozanna [386]3 years ago
7 0

Answer:

1200,1800,2400,3000

5x = 600x

Step-by-step explanation:

The first four terms would be

1200,1800,2400,3000

As we can see that

After 10 weeks of work bob has $1200

After 15 weeks of work, bob has $1,800

So the more the weeks would be, the more the earnings it is

And, the difference of the earnings is

= $1,800 -$1,200

= $600

The equation would be

5x = 600x

So if we assume x = 1

So

5 = 600

if we assume x = 2

So

10 = 1200

and so on

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How many times does 8 go into 135
noname [10]

Answer:

16.875

Step-by-step explanation:

135 divided by 8 equals 16.875

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3 years ago
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Solve the system of linear equations using the
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Answer:

no solution

Step-by-step explanation:

y = -2x + 3

6x + 3y = -3

the substitution method means you plug one equation into the next, because the first equation gives us a solution for y we can go ahead and plug that into y of the second equation

6x + 3(-2x + 3) = -3

6x - 6x + 9 = -3

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8 0
3 years ago
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.

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4 years ago
Add and reduce 2/7+ 16/21=
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Answer:

6/7

Step-by-step explanation:

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The expression simplifies to 3a-21
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