Nine times the sum of a number and eleven is -108. What is the number?

Anna007 [38]

Hi!
The correct answer is B, 1.997 • 10^2

This can be found by simply moving the decimal
place over x times (x represents the exponent). For example, if it were 1.12 • 10^2, you’d move your decimal place over 2 places to the right because it is positive (move left if negative).

7 0

4 years ago

Which point lies on the bisector?

guapka [62]

Answer:

Point Y

Step-by-step explanation:

I have attached a picture of me showing which point lies on the bisector since it says to bisects <C of polygon ABCDE.

Hope this helps, thank you :) !!

8 0

3 years ago

Read 2 more answers

When truckers are on long-haul drives, their driving logs must reflect their average speed. Average speed is the total distance

bekas [8.4K]

a) $v=\frac{d_{tot}}{t_{tot}}=\frac{(3 h)(60 mph)+20 mi}{3 h +t_2}$

The average speed is equal to the ratio between the total distance ( $d_{tot}$ and the total time taken ( $t_{tot}$ ):

$v=\frac{d_{tot}}{t_{tot}}$

the distance travelled by the trucker in the first 3 hour can be written as the time multiplied by the velocity:

$d_1 = (3 h)(60 mph)=180 mi$

So the total distance is

$d_{tot}=d_1 +d_2 = 180 mi+20 mi=200 mi$

The total time is equal to the first 3 hours + the time taken to cover the following 20 miles in the city:

$t_{tot}=3 h +t_2$

So, the equation can be rewritten as:

$v=\frac{d_{tot}}{t_{tot}}=\frac{(3 h)(60 mph)+20 mi}{3 h +t_2}$

b) 0.50 h (half a hour)

Since we know the value of the average speed, $v=57.14 mph$ , we can substitute it into the previous equation to find the value of $t_2$ , the time the trucker drove in the city:

$v=\frac{200 mi}{3h +t_2}\\3h+t_2 = \frac{200 mi}{v}\\t_2 = \frac{200 mi}{v}-3h=\frac{200 mi}{57.14 mph}-3 h=0.50 h$

3 0

3 years ago

Allen�s hummingbird (Selasphorus sasin ) has been studied by zoologist Bill Alther A small group of 15 Allen� s hummingbirds has

Bogdan [553]

Answer:

1) 80% CI: [3.04; 3.26]gr

d= 0.11

2) n= 28 hummingbirds

Step-by-step explanation:

Hello!

The study variable of this experiment is:

X: the weight of a hummingbird. (gr)

And it has a normal distribution, symbolically: X~N(μ;σ²)

And (I hope I got it correctly) its population standard deviation is σ= 0.33

There was a sample of n= 15 hummingbirds taken, its sample mean X[bar]= 3.15 gr

1) You need to construct an 80% Confidence Interval for the population mean of the hummingbird's weight.

Since the study variable has a normal distribution, you can use either the standard normal distribution or the Student's t distribution. Both are useful to estimate the population mean. Since the population standard variance is known, the best choice is the Standard normal.

Z= <u> X[bar] - μ </u>~ N(0;1)

σ/√n

The formula for the interval is:

X[bar] ± $Z_{1- \alpha /2}$ * (σ/√n)

$Z_{1- \alpha /2}= Z_{0.90} = 1.28$

3.15 ± 1.28 * (0.33/√15)

[3.04; 3.26]gr

The margin of error (d) of a confidence interval is hal its amplitude (a)

a= Upper bond - Lower bond

d= (Upper bond - Lower bond)/2

d= $\frac{(3.26-3.04)}{2}$ = 0.11

2) You need to calculate a sample size for a 80% Confidence interval for the average weight of the hummingbirds with a margin of error of d= 0.08

As I said before, the margin of error is half the amplitude of the interval, the formula you use to estimate the population mean has the following structure:

"point estamator" ± "margin of error"

Then the margin of error is:

d= $Z_{1- \alpha /2}$ * (σ/√n)

Now what you have to do is rewrite the formula based on the sample size

d= $Z_{1- \alpha /2}$ * (σ/√n)

$\frac{d}{Z_{1- \alpha /2}}$ = σ/√n

√n * $\frac{d}{Z_{1- \alpha /2}}$ = σ

√n = σ * $\frac{Z_1- \alpha /2}{d}$

n = (σ * $\frac{Z_1- \alpha /2}{d}$ )²

n= (0.33 * $\frac{1.28}{0.08}$ )²

n= 27.8784 ≅ 28 hummingbirds.

I hope it helps!

4 0

3 years ago

There are 23 classes in the fourth grade. Each grade has 21 students. Write a number sentence that will provide a reasonable est

lapo4ka [179]

23 can round to 20 and 21 can also round to 20 so all u do is 20x20=400 so 400 is your estimated answer

6 0

3 years ago