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2 years ago

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The size of the largest angle in a triangle is 3 times the size of the smallest angle. The third angle is 10 more than the small

est angle. Work out the size, in degrees, of each angle in the triangle.

1 answer:

Yakvenalex [24]2 years ago

3 0

Answer:

largest angle=102 smallest angle=34 last angle=44

Step-by-step explanation:

there are 180 degrees in all triangles so 102+44+34=180

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what is the probability of seeing a sample mean for 21 observations less or equal to the sample mean that we observed

kramer

Answer:

$P(x \le 21) = 0.69146$

Step-by-step explanation:

The missing parameters are:

$n = 64$ --- population

$\mu = 20$ --- population mean

$\sigma = 16$ -- population standard deviation

Required

$P(x \le 21)$

First, calculate the sample standard deviation

$\sigma_x = \frac{\sigma}{\sqrt n}$

$\sigma_x = \frac{16}{\sqrt {64}}$

$\sigma_x = \frac{16}{8}$

$\sigma_x = 2$

Next, calculate the sample mean $\bar_x$

$\bar x = \mu$

So:

$\bar x = 20$

So, we have:

$\sigma_x = 2$

$\bar x = 20$

$x = 21$

Calculate the z score

$x = \frac{x - \mu}{\sigma}$

$x = \frac{21 - 20}{2}$

$x = \frac{1}{2}$

$x = 0.50$

So, we have:

$P(x \le 21) = P(z \le 0.50)$

From the z table

$P(z \le 0.50) = 0.69146$

So:

$P(x \le 21) = 0.69146$

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3 years ago

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Step-by-step explanation:

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2 years ago

The position of an object along a vertical line is given by s(t) = −t3 + 3t2 + 7t + 4, where s is measured in feet and t is meas

saw5 [17]

Answer:

The maximum velocity of the object in the time interval [0, 4] is 10 ft/s.

Step-by-step explanation:

Given : The position of an object along a vertical line is given by $s(t) = -t^3+3t^2+7t +4$ , where s is measured in feet and t is measured in seconds.

To find : What is the maximum velocity of the object in the time interval [0, 4]?

Solution :

The velocity is rate of change of distance w.r.t time.

Distance in terms of t is given by,

$s(t) = -t^3+3t^2+7t +4$

Derivate w.r.t. time,

$v(t)=s'(t) = -3t^2+6t+7$

It is a quadratic function so its maximum is at vertex of the function.

The x point of the function is given by,

$x=-\frac{b}{2a}$

Where, a=-3, b=6 and c=7

$t=-\frac{6}{2(-3)}$

$t=-\frac{6}{-6}$

$t=1$

As 1 lie between interval [0,4]

Substitute t=1 in the function,

$v(t)= -3(1)^2+6(1)+7$

$v(t)= -3+6+7$

$v(1)=10$

Th maximum velocity is 10 ft/s.

Therefore, the maximum velocity of the object in the time interval [0, 4] is 10 ft/s.

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