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KIM [24]
3 years ago
6

Find the greatest common factor of 14 and 28​

Mathematics
1 answer:
Over [174]3 years ago
8 0
The greatest common factor of 14 and 28 is 14
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The measures of the angles of a triangle are shown in the figure below. Solve for x.
ArbitrLikvidat [17]

Answer:

14

Step-by-step explanation:

5x-3+ x+9 + 90= 180

6x+ 96=180

6x=180-96

6x=84

x=84/6

x=14

3 0
3 years ago
If arc BC is 42° and angle 5 is 32°, what is the measure of arc DE?
pychu [463]

Answer:

22

Step-by-step explanation:

6 0
3 years ago
In an election, the population consists of the people who voted. Although there is overall data on how the population voted, the
andrew-mc [135]

Answer:

Answer:

P(A) = 0.39

Step-by-step explanation:

We are given;

P(W|A) = 0.7

P(W|A^c ) = 0.3

We are told that 60% of the respondents said they voted for A. Thus;

P(A|W) = 60% = 0.6

Now, using the principle of drawing lots, we can be able to find the probability of the event that they are willing to participate in the exit poll which is P(W).

Thus;

P(W) = [P(W|A) × P(A)] +[P(W∣A^c) × P(A^c)]

Now, P(A^c) can be expressed as 1 - P(A)

Thus, we now have;

P(W) = [P(W|A) × P(A)] + [P(W∣A^c) × (1 - P(A)]

Plugging in the relevant values gives;

P(W) = 0.7P(A) + 0.3(1 - P(A))

P(W) = 0.7P(A) + 0.3 - 0.3P(A)

P(W) = 0.3 + 0.4P(A)

Now,using Baye's theorem, we can find an expression for P(A|W)

Thus;

P(A|W) = [P(A ∩ W)]/P(W)

This can be further expressed as;

P(A|W) = [P(A) × P(W|A)]/P(W)

Plugging in relevant values, we have;

0.6 = 0.7P(A)/(0.3 + 0.4P(A))

Cross multiply to get;

0.6(0.3 + 0.4P(A)) = 0.7P(A)

0.18 + 0.24P(A) = 0.7P(A)

0.18 = 0.7P(A) - 0.24P(A)

0.46P(A) = 0.18

P(A) = 0.18/0.46

P(A) = 0.39

Step-by-step explanation:

braniest

4 0
3 years ago
What is the value of n - 5, when n = 8?<br> A) 3 <br> B) 4 <br> C) 12 <br> D) 13
evablogger [386]
You plug 8 into where n is so it's 8-5
8-5= 3 so the answer is A
8 0
3 years ago
Read 2 more answers
Sin(x2 + y2) da r , where r is the region in the first quadrant between the circles with center the origin and radii 2 and 5
sweet [91]
Converting to polar coordinates gives

\displaystyle\iint_R\sin(x^2+y^2)\,\mathrm dA=\int_{\theta=0}^{\theta=\pi/2}\int_{r=2}^{r=5}\sin(r^2)r\,\mathrm dr\,\mathrm d\theta
=\displaystyle\pi\int_{r=2}^{r=5}2r\sin(r^2)\,\mathrm dr
=\displaystyle\pi\int_{r^2=4}^{r^2=25}\sin(r^2)\,\mathrm d(r^2)
=-\cos(r^2)\bigg|_{r^2=4}^{r^2=25}
=\pi(\cos4-\cos25)
3 0
3 years ago
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