Please help me out with this.

Mathematics

1 answer:

xenn [34]3 years ago

5 0

Answer:

80 rooms

Step-by-step explanation:

since 45% of the rooms are booked, and 36 rooms are booked, then this means that 36 = 45% of x.

then, you we have 45% × x = 36

or $\frac{45}{100}$ × x = 36

Multiplying both sides by 100 and dividing both sides by 45,

we have x = 36 × $\frac{100}{45}$

= $\frac{3600}{45}$

x = 80

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PLEASE HURRY THIS IS DUE IN 20MIN

OverLord2011 [107]

This is not my answer I found it from another brainly

From the beginning we know we are dealing with two different rates. Distance is equal to (Speed X Time). We have two different speeds though. We will call the min speed = x and the max speed = y. With those two ideas in place we can make two equations for both Tony and Rae.

Tony:

2x+3.5y=355

Rae:

2x+3y=320

Now you can solve the equations by subtracting Rae's equation from Tony's

.5y=35 solve for y and you get 70

Now plug that y value back into either equation and you can solve for x. I'll use Rae's distance:

2x+3(70)=320

x=55

6 0

3 years ago

Life tests on a helicopter rotor bearing give a population mean value of 2500 hours and a population standard deviation of 135 h

stira [4]

Answer:

The percent of the parts are expected to fail before the 2100 hours is 0.15.

Step-by-step explanation:

Given :Life tests on a helicopter rotor bearing give a population mean value of 2500 hours and a population standard deviation of 135 hours.

To Find : If the specification requires that the bearing lasts at least 2100 hours, what percent of the parts are expected to fail before the 2100 hours?.

Solution:

We will use z score formula

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

Mean value = $\mu = 2500$

Standard deviation = $\sigma = 135$

We are supposed to find If the specification requires that the bearing lasts at least 2100 hours, what percent of the parts are expected to fail before the 2100 hours?

So we are supposed to find P(z<2100)

so, x = 2100

Substitute the values in the formula

$z=\frac{2100-2500}{135}$