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2 years ago

If m/CHF = 106°, then what is m/FHD?

Mathematics

1 answer:

Tom [10]2 years ago

7 0

m<CHF and m<FHD are supplementary

$\\ \rm\Rightarrow 106+\angle{FHD}=180$

$\\ \rm\Rightarrow m\angle{FHD}=180-106$

$\\ \rm\Rightarrow m\angle{FHD}=74$

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Which is the graph of −2x+y+4<9? HELP PLEASE!

Sergio039 [100]

Answer:

b kno its not right but crown me lol

Step-by-step explanation:

4 0

3 years ago

The members of the Student Activity Council on your campus are meeting to select three speakers for a month-long event celebrat

Komok [63]

The number of different ways that three speakers can be selected is 10 ways

Given the names of choice to be Ben, Will, Stewart, Hilary, and Kate. This means that we have a total of 5 name choices.

If the members of students activities are to select three speakers among these people, the number of ways this can be done is by using the combination rule as shown;

$nCr=\frac{n!}{(n-r)!r!}$

From the question, n = 5 and r = 3. On substituting

$5C3=\frac{5!}{(5-3)!3!}\\5C3=\frac{5!}{2!3!}\\5C3=\frac{5 \times 4 \times 3!}{2 \times 3!} \\5C3=\frac{20}{2}\\5C3 = 10 ways$

Hence the number of different ways that three speakers can be selected is 10 ways.

Learn more here: brainly.com/question/24145745

7 0

3 years ago

Use the product property of roots to choose the expression equivalent to 3√5x*3√25x^2?

nignag [31]

Answer: $\sqrt[3]{125x^3}$ .

Step-by-step explanation: Given radical expression

$\sqrt[3]{5x} \times \sqrt[3]{25x^2}$ .

According to the product property of roots.

$\sqrt[n]{a} \times \sqrt[n]{b} = \sqrt[n]{a \times b}$

On applying above rule, we get

$\sqrt[3]{5x} \times \sqrt[3]{25x^2} = \sqrt[3]{5x \times 25x^2}$

5 × 25 = 125 and

$x \times x^2 = x^3$

Therefore,

$\sqrt[3]{5x \times 25x^2}= \sqrt[3]{125x^3}$

<h3>So, the correct option would be second option $\sqrt[3]{125x^3}$ .</h3>

3 0

3 years ago

Read 2 more answers

J= -8c^2 - 4cd - d^2 k= -3c^2 +7cd - 3d^2 j+k =

irina [24]

J=-8c²-4cd-d²
k=-3c²+7cd-3d²
j+k=
-8c²-4cd-d²-3c²+7cd-3d²
-11c²-4d²+3cd

6 0

4 years ago

The manager of a computer retails store is concerned that his suppliers have been giving him laptop computers with lower than av

34kurt

Answer:

Probability that the 50 randomly selected laptops will have a mean replacement time of 3.1 years or less is 0.0092.

Yes. The probability of this data is unlikely to have occurred by chance alone.

Step-by-step explanation:

We are given that the replacement times for the model laptop of concern are normally distributed with a mean of 3.3 years and a standard deviation of 0.6 years.

He then randomly selects records on 50 laptops sold in the past and finds that the mean replacement time is 3.1 years.

Let M = sample mean replacement time

The z-score probability distribution for sample mean is given by;

Z = $\frac{ M-\mu}{\frac{\sigma}{\sqrt{n} } }} }$ ~ N(0,1)

where, $\mu$ = population mean replacement time = 3.3 years

$\sigma$ = standard deviation = 0.6 years

n = sample of laptops = 50

The Z-score measures how many standard deviations the measure is away from the mean. After finding the Z-score, we look at the z-score table and find the p-value (area) associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X.

Now, Probability that the 50 randomly selected laptops will have a mean replacement time of 3.1 years or less is given by = P(M $\leq$ 3.1 years)

P(M $\leq$ 3.1 years) = P( $\frac{ M-\mu}{\frac{\sigma}{\sqrt{n} } }} }$ $\leq$ $\frac{ 3.1-3.3}{\frac{0.6}{\sqrt{50} } }} }$ ) = P(Z $\leq$ -2.357) = 1 - P(Z $\leq$ 2.357)

= 1 - 0.99078 = 0.0092 or 0.92%

So, in the z table the P(Z $\leq$ x) or P(Z < x) is given. So, the above probability is calculated by looking at the value of x = 2.357 in the z table which will lie between x = 2.35 and x = 2.36 which has an area of 0.99078.

Hence, the required probability is 0.0092 or 0.92%.

Now, based on the result above; Yes, the computer store has been given laptops of lower than average quality because the probability of this data is unlikely to have occurred by chance alone as the probability of happening the given event is very low as 0.92%.

8 0

4 years ago