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sammy [17]
3 years ago
7

FIND THE MISSING LENGTH!!! URGENT!!!

Mathematics
1 answer:
Pani-rosa [81]3 years ago
5 0

Answer:

4

Step-by-step explanation:

Triangle BUC is 3 times smaller than Triangle SUT. This means that you would divide 12/3 and get 4 as the missing side.

Hope it helps!

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What’s the work and answer for this? Someone please tell me
Yuliya22 [10]

Answer:

3

Step-by-step explanation:

3 0
3 years ago
Evaluate 1/3m-1-1/2n when m=21 and n=12​
siniylev [52]
The answer is 0, 1/3 of 21 is 7, 7-1=6, 1/2 of 12= 6, and 6-6=0. Hope this helps!
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.

<u><em /></u>

<u><em>Let </em></u>\mu<u><em> = average mg of mercury in compact fluorescent bulbs.</em></u>

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 <u>One-sample z test</u> <u>statistics</u> 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, <em><u>test statistics</u></em>  =  \frac{3.59-3.50}{\frac{0.18}{\sqrt{25} } }

                               =  2.50

The value of z test statistics is 2.50.

<u>Now, P-value of the test statistics is given by;</u>

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

                                             = 1 - 0.9938 = 0.0062

<em />

<em>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 </em><em><u>we reject our null hypothesis</u></em><em>.</em>

Therefore, we conclude that the average mg of mercury in compact fluorescent bulbs is more than 3.50 mg.

6 0
3 years ago
Money Flow  The rate of a continuous money flow starts at $1000 and increases exponentially at 5% per year for 4 years. Find the
77julia77 [94]

Answer:

Present value =  $4,122.4

Accumulated amount = $4,742

Step-by-step explanation:

Data provided in the question:

Amount at the Start of money flow = $1,000

Increase in amount is exponentially at the rate of 5% per year

Time = 4 years

Interest rate = 3.5%  compounded continuously

Now,

Accumulated Value of the money flow = 1000e^{0.05t}

The present value of the money flow = \int\limits^4_0 {1000e^{0.05t}(e^{-0.035t})} \, dt

= 1000\int\limits^4_0 {e^{0.015t}} \, dt

= 1000\left [\frac{e^{0.015t}}{0.015} \right ]_0^4

= 1000\times\left [\frac{e^{0.015(4)}}{0.015} -\frac{e^{0.015(0)}}{0.015} \right]

= 1000 × [70.7891 - 66.6667]

= $4,122.4

Accumulated interest = e^{rt}\int\limits^4_0 {1000e^{0.05t}(e^{-0.035t}} \, dt

= e^{0.035\times4}\times4,122.4

= $4,742

8 0
3 years ago
Log(x-2)-log (2x-3) = log2​
ivanzaharov [21]

Answer:

Use the property of condensation of logarithmic function

Given,

7 0
3 years ago
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