All categories

3 years ago

NEED ASAP WILL MARK BRAINLIEST!!!!

Mathematics

1 answer:

Paha777 [63]3 years ago

3 0

Answer

(40+y*0.02)+0.07

0.02y+40.07

Step-by-step explanation:

You might be interested in

(04.02 MC) Solve the system of equations. 5x + y = 9 3x + 2y = 4

goblinko [34]

Answer:

(2,-1)

Step-by-step explanation:

Multiply first equation by 2:

10x + 2y = 18

Subtract the second equation from this so that the 2y terms cancel:

10x - 3x = 18 - 4

7x = 14

x = 2

Plug into first equation to find y:

5(2) + y = 9

10 + y = 9

y = -1

The answer is (2,-1)

7 0

4 years ago

Read 2 more answers

"You measure 34 dogs' weights, and find they have a mean weight of 67 ounces. Assume the population standard deviation is 13.5 o

liraira [26]

Answer:

The 95% confidence interval for the true population mean dog weight is between 62.46 ounces and 71.54 ounces.

Step-by-step explanation:

We have that to find our $\alpha$ level, that is the subtraction of 1 by the confidence interval divided by 2. So:

$\alpha = \frac{1-0.95}{2} = 0.025$

Now, we have to find z in the Ztable as such z has a pvalue of $1-\alpha$ .

So it is z with a pvalue of $1-0.025 = 0.975$ , so $z = 1.96$

Now, find M as such

$M = z*\frac{\sigma}{\sqrt{n}}$

In which $\sigma$ is the standard deviation of the population and n is the size of the sample.

$M = 1.96*\frac{13.5}{\sqrt{34}} = 4.54$

The lower end of the interval is the sample mean subtracted by M. So it is 67 - 4.54 = 62.46 ounches.

The upper end of the interval is the sample mean added to M. So it is 67 + 4.54 = 71.54 ounces.

The 95% confidence interval for the true population mean dog weight is between 62.46 ounces and 71.54 ounces.

7 0

4 years ago

Raymond has taken a position as the navigator on a speed boat racing team. The team entered a race that starts at Daytona Beach,

erik [133]

Using the rules of cosines, the measure of the turn between the second and third legs of the race is given by

$\cos^{-1}\left( \frac{180^2+160^2-140^2}{2\times180\times160} \right)=\cos^{-1}\left( \frac{32,400+26,600-19,600}{57,600} \right) \\ \\ =\cos^{-1}\left( \frac{39,400}{57,600} \right)=\cos^{-1}(0.6840)\approx47^o$