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Tcecarenko [31]
2 years ago
9

7) 4b2 + 8b + 7 = 4 What the answer

Mathematics
2 answers:
Anit [1.1K]2 years ago
8 0

4b² + 8b + 3 = 0

4b² + 2b + 6b + 3 = 0

2b(2b + 1) + 3(2b + 1) = 0

(2b + 3)(2b + 1) = 0

b = -1/2 or b = -3/2

san4es73 [151]2 years ago
7 0
B= -1/2 or b= -3/2

1. Subtract 4 from both sides.
2. Factor left side of equation.
3. Set factors equal to 0.
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bonufazy [111]

Answer:

$10

Step-by-step explanation:

The question made is How much did I have? This question is about the amount of money that you had before receiving the money from your mom, dad and aunt and uncle. This means that the answer is the $3 you had plus the other $7 you said you had. So, $3+$7=$10. You had $10.

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A particular brand of tires claims that its deluxe tire averages at least 50,000 miles before it needs to be replaced. From past
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Answer:

We conclude that deluxe tire averages less than 50,000 miles before it needs to be replaced which means that the claim is not supported.

Step-by-step explanation:

We are given that a particular brand of tires claims that its deluxe tire averages at least 50,000 miles before it needs to be replaced. From past studies of this tire, the standard deviation is known to be 8000.

From the 28 tires surveyed, the mean lifespan was 46,500 miles with a standard deviation of 9800 miles.

<u><em>Let </em></u>\mu<u><em> = average miles for deluxe tires</em></u>

So, Null Hypothesis, H_0 : \mu \geq 50,000 miles   {means that deluxe tire averages at least 50,000 miles before it needs to be replaced}

Alternate Hypothesis, H_A : \mu < 50,000 miles    {means that deluxe tire averages less than 50,000 miles before it needs to be replaced}

The test statistics that will be used here is <u>One-sample z test statistics</u> as we know about population standard deviation;

                                  T.S.  = \frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }  ~ N(0,1)

where, \bar X = sample mean lifespan = 46,500 miles

            \sigma = population standard deviation = 8000 miles

            n = sample of tires = 28

So, <u><em>test statistics</em></u>  =  \frac{46,500-50,000}{\frac{8000}{\sqrt{28} } }

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The value of the test statistics is -2.315.

Now at 5% significance level, the z table gives critical value of -1.6449 for left-tailed test. Since our test statistics is less than the critical value of z as -2.315 < -1.6449, so we have sufficient evidence to reject our null hypothesis as it will fall in the rejection region due to which we reject our null hypothesis.

Therefore, we conclude that deluxe tire averages less than 50,000 miles before it needs to be replaced which means that the claim is not supported.

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