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3 years ago

Please help I need to have this in by tonight

Mathematics

2 answers:

lbvjy [14]3 years ago

7 0

The answer is -0.045y

Daniel [21]3 years ago

4 0

Hi, I’ll gladly be happy to help, but could you type out the problem instead because the photo is too blurry for me to see.

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Choose the two rectangular prisms that have a volume of 36cm^3?

oee [108]

Answer:

Correct choices: B, D

Step-by-step explanation:

The Volume of a Rectangular Prism

Given a rectangular prism of dimension W, L, and H, the volume is calculated as:

V = WLH

Each small cube of the figure has a volume of 1 cube centimeter. We need to calculate the volume of each solid and determine which ones have a volume of 36 cube cm.

A: V = 5*3*4 = 60 cube cm

B: V = 3*3*4 = 36 cube cm

C: V = 3*4*2 = 24 cube cm

D: V = 6*2*3 = 36 cube cm

E: V = 2*4*2 = 16 cube cm

Correct choices: B, D

8 0

2 years ago

A baker is using a cookie recipe that calls for 2 1/4 cups of flower to yield 36 cookies. How many flour will the baker need to

Kryger [21]

90/36 is 2.5, so the Baker needs to make 2.5 batches of cookies

2.5 batches of cookies x 2.25 cups of flour

is 5.625 cups of flour

which is 5 5/8 cups of flour

6 0

2 years ago

In place value the Relationship between 2,000 and 20 is how many times

Katen [24]

Answer:

100

Step-by-step explanation:

20 times 100 equals 2000

6 0

3 years ago

Find the volume of the solid under the surface z = 5x + 9y 2 and above the region bounded by x = y 2 and x = y 3.

Yanka [14]

Call the region in the $x$ - $y$ plane, bounded by $x=y^2$ and $x=y^3$ , $\mathcal D$ . Then the volume under the given surface is

$\displaystyle\iint_{\mathcal D}(5x+9y)\,\mathrm dA=\int_{y=0}^{y=1}\int_{x=y^3}^{x=y^2}(5x+9y)\,\mathrm dx\,\mathrm dy=\frac{83}{140}$

8 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