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3 years ago

K^2+8k-9=0 can you help me solve with showing work

2 answers:

7 0

Using the PSN method:

P = -9
S = 8
So N = 9,-1

Using 9 and -1 in the original equation gives:

$k^2+9k-k-9=0
\\k(k+9)-1(k+9)=0
\\(k-1)(k+9)

\\\\k=1,-9$

erastova [34]3 years ago

5 0

K² + 8k - 9 = 0
k = -8 +/- √(8² - 4(10)(-9))
 2(1)
k = -8 +/- √(64 + 360)
 2
k = -8 +/- √(424)
 2
k = -8 +/- 20.59126028
 2
k = -8 + 20.59126028 k = -8 - 20.59126028
 2 2
k = 12.5912608 k = -20.5912608
 2 2
k = 6.2956304 k = -10.2956304

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Answer:

We conclude that the mean resting pulse rate for men is 72 beats per minute.

Step-by-step explanation:

We are given that he mean resting pulse rate for men is 72 beats per minute. Assume that the standard deviation of the resting pulse rates of all men who work out at Mitch's Gym is known to be 2.6 beats per minute.

A simple random sample of men who regularly work out at Mitch's Gym is obtained and their resting pulse rates (in beats per minute) are listed below;

87, 89, 69, 63, 70, 65, 88, 84, 58, 53, 66, 70.

Let $\mu$ = mean resting pulse rate for men.

SO, Null Hypothesis, $H_0$ : $\mu$ = 72 beats/minute {means that the mean resting pulse rate for men is 72 beats per minute}

Alternate Hypothesis, $H_A$ : $\mu$ < 72 beats/minute {means that the mean resting pulse rate for men is less than 72 beats per minute}

The test statistics that would be used here One-sample z-test statistics 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 mean resting pulse rate = $\frac{\sum X}{n}$

= $\frac{87+ 89+ 69 +63 +70 +65+ 88+ 84+ 58+ 53+ 66+ 70}{12}$ = 71.83 beats/minute

$\sigma$ = population standard deviation = 2.6 beats per minute

n = sample of men = 12

So, the test statistics = $\frac{71.83 - 72}{\frac{2.6}{\sqrt{12} } }$

= -0.23

The value of z test statistics is -0.23.

Also, P-value of the test statistics is given by;

P(Z < -0.23) = 1 - P(Z $\leq$ 0.23)

= 1 - 0.59095 = 0.40905

Now, at 0.05 significance level the z table gives critical value of -1.645 for left-tailed test.

Since our test statistic is more than the critical value of z as -0.23 > -1.645, so we have insufficient evidence to reject our null hypothesis as it will not fall in the rejection region due to which we fail to reject our null hypothesis.

Therefore, we conclude that the mean resting pulse rate for men is 72 beats per minute.

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