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klio [65]
3 years ago
11

Determine the solution to each linear system by using the substitution method.

Mathematics
1 answer:
lana66690 [7]3 years ago
3 0
A. 2x+3×5=34
    2x+15=34
   2×9.5=19 
     2×9.5+15=34

x=9.5
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What is equivalent to 3:9
Gnesinka [82]

Answer:

1:3

Lowest possible simplification.

4 0
3 years ago
Read 2 more answers
14-(x+5)=3(x+9)-10 please help quick and show work!
avanturin [10]

Answer:

x=-2

Step-by-step explanation:

14-(x+5)=3(x+9)-10

14-x-5=3x+27-10

14-5-x=3x+17

     +x  +x

9=4x+17

-17    -17

4x=-8

4x/4=-8/4

x=-2

6 0
2 years ago
Simplify:<br> 8(v+w) - 7(v+2w)
Liula [17]
First you need to distribute. when you do this you get (8v+8w)-(7v+14w) then you get like terms together, but don't forget that everything in (7v+14w) is negative. when you group like terms together you get (8v-7v)+(8w-14w) simplify and you get v-6w
4 0
3 years ago
Read 2 more answers
Eight more than the product of four and a number is twice the sum of the number and six. Write an equation that can be used to s
masha68 [24]
4n+8=2(n+6)
Now isolate the n on one side of the equation:
4n+8=2n+12(distribution of the 2)
4n=2n+4(combine like terms)
2n=4(combine like variables)
N=2(isolate the n)

The answer is 2
Good luck!
3 0
2 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.

6 0
3 years ago
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