Write in terms of i. Simplify as much as possible. <br><br> -√-45

there are 400 students and 4/5 have purchased tickets there needs to be 2 chaperones for every 75 students how many chaperones a

Vesna [10]

Answer:

5 chaperones are needed.

Step-by-step explanation:

400 x 4/5 = 320

320 ÷ 75 = 4.3

3 0

2 years ago

I need help with question 18 PLS!!!!

dolphi86 [110]

Answer:

162

Step-by-step explanation:

( -a ) ( b ) ( -a + b )

-6 ( 3 ) ( -6 + -3 )

-6 ( 3 ) ( -9 )

- 18 ( -9 )

162

6 0

2 years ago

A, B and C can do a piece of work in 20, 30 and 60 days respectively. In how many days can A do the work if he is assisted by B

Goshia [24]

A, B and C working together will finish same piece of work in 10 days

A can do a piece of work in 20 days. That is A is doing the work at a rate of 1 / 20. Therefore

B = 1 / 30

C = 1 / 60

A, B and C working togehter their rate will be added together =

A + B + C = 1 / 20 + 1 / 30 + 1 / 60 = 1 / 10

so they can work together to finish same piece of work in 10 days

4 0

1 year ago

If sin 0=3/5, then cos 0=

vaieri [72.5K]

Answer:

$\frac{4}{5}$

Step-by-step explanation:

Given

sinΘ = $\frac{3}{5}$ = $\frac{opposite}{hypotenuse}$

Then the hypotenuse = 5 and one leg of the right triangle is 3

Using Pythagoras' identity to find the second leg (x)

x² + 3² = 5²

x² + 9 = 25 ( subtract 9 from both sides )

x² = 16 ⇒ x = 4

cosΘ = $\frac{adjacent}{hypotenuse}$ = $\frac{4}{5}$

7 0

2 years ago

Construct the confidence interval for the population mean mu. c = 0.90, x = 16.9, s = 9.0, and n = 45. A 90% confidence inte

Georgia [21]

Answer:

The 90% confidence interval for population mean is $14.7 < \mu < 19.1$

Step-by-step explanation:

From the question we are told that

The sample mean is $\= x = 16.9$

The confidence level is $C = 0.90$

The sample size is $n = 45$

The standard deviation

Now given that the confidence level is 0.90 the level of significance is mathematically evaluated as

$\alpha = 1-0.90$

$\alpha = 0.10$

Next we obtain the critical value of $\frac{\alpha }{2}$ from the standardized normal distribution table. The values is $Z_{\frac{\alpha }{2} } = 1.645$

The reason we are obtaining critical values for $\frac{\alpha }{2}$ instead of that of $\alpha$ is because $\alpha$ represents the area under the normal curve where the confidence level 1 - $\alpha$ (90%) did not cover which include both the left and right tail while $\frac{\alpha }{2}$ is just considering the area of one tail which is what we required calculate the margin of error