A system of two linear equations and three unknowns can be solved by:

Jenna and her mom are addressing envelopes for Jenna's wedding invitations. So far, Jenna has addressed 2/9 of the envelopes, an

frutty [35]

I can’t tell if you want fraction or decimal so ima say 4/9

5 0

3 years ago

Read 2 more answers

Express 2x^2+8x+3 in the form 2(x+p)^2+q

e-lub [12.9K]

Answer:

2(x + 2)² - 5

Step-by-step explanation:

Given

2x² + 8x + 3

To obtain the required form use the method of completing the square.

The coefficient of the x² term must be 1, thus factor 2 out of 2x² + 8x

= 2(x² + 4x) + 3

add/subtract ( half the coefficient of the x- term)² to x² + 4x

= 2(x² + 2(2)x + 4 - 4) + 3

= 2(x + 2)² - 8 + 3

= 2(x + 2)² - 5

with p = 2 and q = - 5

4 0

3 years 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

2 years ago

What are the possible outcomes?

vesna_86 [32]

Answer:

A= X,B

B= L,G

C= M,W

The total number of combinations is 12

6 0

3 years ago

An isosceles triangle has side lengths of 15.6 units and 20.33 units. What is the greatest possible value of the perimeter of th

ArbitrLikvidat [17]

Answer:

56.26

Step-by-step explanation:

An isosceles triangle has two sides that are the same length, and one side that is the same length. We have two sizes, so two find the greatest possible perimeter we must assume that the greater size is the value for the two sides, and the smaller size the value for the single side. In other words:

$x = 15.6 + 2(20.33)$

$x = 56.26$

6 0

1 year ago