All categories

3 years ago

The Siberian tiger can grow up to 10 4/5 feet long. Express this length as decimal

Mathematics

1 answer:

MrRissso [65]3 years ago

6 0

The answer is 10.8. You turn 10 4/5 into 54/5. Then you divide 54 by 5.

You might be interested in

runben sees 14 wheels on a total of 6 bicycles and tricycles how many bicycles and tricycles are there

olganol [36]

Answer:

B = 4

T = 2

Step-by-step explanation:

First, figure out the equation.

We know that bicycles have 2 wheels, that there are 3 on a tricycle, and there are a total of 14 wheels and a total of 6 bicycles and tricycles.

Let b stand for bicycles and t stand for tricycles:

14 = 2b + 3t

6 = b + t

We can figure out the amount of either by rearranging the second equation to isolate one variable. I will solve it in two ways

In the first way, I will solve for b

(-t) 6 = b + t (-t)

6 - t = b

Plug this into the first equation and solve for remaining variable

14 = 2(6 - t) +3t

14 = 12 - 2t + 3t

14 = 12 +t

-12 -12

2 = t

6 - 2 = b

b = 4

The second way was to solve for t first

(-b) 6 = b + t (-b)

6 - b = t

14 = 2b + 3(6 - b)

14 = 2b + 18 - 3b

(-18) 14 = 18 -b (-18)

-4/-1 = -b/-1

b = 4

6 - 4 = t

t = 2

It doesn't matter which way you go, they both give you the exact same answer.

Sooo, recap!

1) write equations

2) switch the easier of the two to isolate one variable

3) substitute to find other variable (x2)

4) Find answers! =D

Hope this helps!

8 0

2 years ago

A Rasmussen Reports survey of 1,000 US adults found that 42% believe raising the minimum wage will help the economy. Construct a

vitfil [10]

Answer:

(0.3798, 0.4602)

Step-by-step explanation:

Let p be the true proportion of US adults who believe raising the minimum wage will help the economy. A point estimate for p is $\hat{p}=0.42$ and a good aproximation (because we have a large sample) for the standard deviation of $\hat{p}$ is $\sqrt{\hat{p}(1-\hat{p})/n}=\sqrt{0.42(0.58)/1000}=0.0156$ . Therefore, a 99% confidence interval for p is given by $\hat{p}\pm z_{\alpha/2}\sqrt{\frac{\hat{p}(1-\hat{p})}{n}}$ where $\alpha=0.01$ and $z_{\alpha/2}$ is the $\alpha/2$ th quantile of the standard normal distribution, so we have, $0.42\pm z_{0.005}0.0156$ , i.e., $0.42\pm(-2.5758)(0.0156)$ or equivalently (0.3798, 0.4602).