Find the 9th term of the sequence described by: A(n)=7+(n-1)(3)

Find the value of x. Express your answer in simplest radical form

klemol [59]

Since x=x, this is an isosceles right triangle. By the Pythagorean Theorem:

h^2=a^2+b^2 (the hypotenuse squared is equal to the sum of the squared sides)

5^2=x^2+x^2

25=2x^2

2x^2=25

x^2=25/2

x=√(25/2)

x=5/√2 now if we rationalize the denominator...

x=(5√2)/(√2√2)

x=(5√2)/2

7 0

3 years ago

Help asap please,i don’t get the question

mario62 [17]

Answer:

see below

Step-by-step explanation:

I do not know what the 'given expression' is, bu the three UN-highlighted expressions in each question 1 and 2 are equivalent

Only the highlighted one is not equiv to the other three

3 0

1 year ago

GreenBeam Ltd. claims that its compact fluorescent bulbs average no more than 3.50 mg of mercury. A sample of 25 bulbs shows a m

Harrizon [31]

Answer:

(a) Null Hypothesis, $H_0$ : $\mu \leq$ 3.50 mg

Alternate Hypothesis, $H_A$ : $\mu$ > 3.50 mg

(b) The value of z test statistics is 2.50.

(c) We conclude that the average mg of mercury in compact fluorescent bulbs is more than 3.50 mg.

(d) The p-value is 0.0062.

Step-by-step explanation:

We are given that Green Beam Ltd. claims that its compact fluorescent bulbs average no more than 3.50 mg of mercury. A sample of 25 bulbs shows a mean of 3.59 mg of mercury. Assuming a known standard deviation of 0.18 mg.

Let $\mu$ = average mg of mercury in compact fluorescent bulbs.

So, Null Hypothesis, $H_0$ : $\mu \leq$ 3.50 mg {means that the average mg of mercury in compact fluorescent bulbs is no more than 3.50 mg}

Alternate Hypothesis, $H_A$ : $\mu$ > 3.50 mg {means that the average mg of mercury in compact fluorescent bulbs is more than 3.50 mg}

The test statistics that would be used here One-sample z test statistics as we know about the population standard deviation;

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

where, $\bar X$ = sample mean mg of mercury = 3.59

$\sigma$ = population standard deviation = 0.18 mg

n = sample of bulbs = 25

So, test statistics = $\frac{3.59-3.50}{\frac{0.18}{\sqrt{25} } }$

= 2.50

The value of z test statistics is 2.50.

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

P-value = P(Z > 2.50) = 1 - P(Z $\leq$ 2.50)

= 1 - 0.9938 = 0.0062

Now, at 0.01 significance level the z table gives critical value of 2.3263 for right-tailed test. Since our test statistics is more than the critical values of z, 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 the average mg of mercury in compact fluorescent bulbs is more than 3.50 mg.

6 0

3 years ago

Help please! worth a lot of points! Screenshot included

vlada-n [284]

Answer:

x=44444444444444444444444444444444

Step-by-step explanation:

8 0

2 years ago

Read 2 more answers

Which one equals 16 if you set it up like this? Please Halp

Sveta_85 [38]

The first one equals 16. The second one equals 0.0625

6 0

3 years ago