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Nady [450]
2 years ago
3

Convert the angle to radians. Round to three decimal places.

Mathematics
1 answer:
xxMikexx [17]2 years ago
7 0

Given:

52^{\circ}\text{Angle in radian=52}\times\frac{\pi}{180}\text{Angle in radian=}52\times\frac{3.14}{180}\text{Angle in radian=}0.90752^{\circ}=0.907\text{ radians}

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-2 1/4 + -4 1/2<br><br>(Please answer as fast as possible)
lianna [129]

Answer: -6 3/4

Short explained no step by step

3 0
3 years ago
Once a fire is reported to a fire insurance company, the company makes an initial estimate, X, of the amount it will pay to the
Olenka [21]

Answer:

The probability that the final settlement amount is between 1 and 3 given that the initial claim is 2 = (2/9) = 0.2222

Step-by-step explanation:

The complete question is presented in the attached image to this solution

The joint probability distribution is given as

f(x, y) = {2/[x²(x - 1)} × y^-[(2x-1)/(x-1)] for x>1 And y>1

Given that the initial claim estiamted by the comapny is 2, determine the probability that the final settlement amount is between 1 and 3.

That is, x = 2, and y ranges from 1 to 3

Inserting x = 2 into the expression, we obtain

f(y) = (1/2) × y⁻³ = (y⁻³/2)

The required probability would then be

P(1 < y ≤ 3) = ∫³₁ f(y) dy

= ∫³₁ (y⁻³/2) dy

= [y⁻²/-4]³₁

= [3⁻²/-4] - [1⁻²/-4]

= (-1/36) - (-1/4)

= (1/4) - (1/36)

= (8/36)

= (2/9) = 0.2222

Hope this Helps!!!

4 0
4 years ago
There is a box with an area of 112 sq inches and a height of 3.5 inches. What is the width of the box?
Rashid [163]

Answer:

32 inches

Step-by-step explanation:

Because height times width equals area,

we can divide 112 by 3.5 to get the width

So it is 32 inches

8 0
4 years ago
The marketing director of a large department store wants to estimate the average number of customers who enter the store every f
Dahasolnce [82]

Answer:

36.5674\leq x'\leq61.4326

Step-by-step explanation:

If we assume that the number of arrivals is normally distributed and we don't know the population standard deviation, we can calculated a 95% confidence interval to estimate the mean value as:

x-t_{\alpha /2}\frac{s}{\sqrt{n} } \leq x'\leq x-t_{\alpha /2}\frac{s}{\sqrt{n} }

where x' is the population mean value, x is the sample mean value, s is the sample standard deviation, n is the size of the sample, \alpha is equal to 0.05 (it is calculated as: 1 - 0.95) and  t_{\alpha /2} is the t value with n-1 degrees of freedom that let a probability of \alpha/2 on the right tail.

So, replacing the mean of the sample by 49, the standard deviation of the sample by 17.38, n by 10 and t_{\alpha /2} by 2.2621 we get:

49-2.2621\frac{17.38}{\sqrt{10} } \leq x'\leq 49+2.2621\frac{17.38}{\sqrt{10} }\\49-12.4326\leq x'\leq 49+12.4326\\36.5674\leq x'\leq61.4326

Finally, the interval values that she get is:

36.5674\leq x'\leq61.4326

8 0
3 years ago
Find the x-intercept of the following line. y=5x+20
Pavlova-9 [17]

Answer:

(-4, 0)

Step-by-step explanation: I recommend using desmos graphing calculator for problems like these. It will graph your problem and you can see where it connects. Hope this helps!

3 0
3 years ago
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