Lilli jogs 1/2 mile in 1/10 hour.

..1+4=5. 2+5=12. 3+6=21. 8+11=40 or this could be 96 which is right and why

andre [41]

Multiply the numbers being added then add the first number. 4X1=4, 4+1=5.
2X5=10, 10+2=12. 11X8=88, 88+8=96. I'm not sure how to get 40, though.
Hope I could help.

7 0

3 years ago

What's the answer to this math problem?

m_a_m_a [10]

Answer:

take the numbers and do the equation steps then fill the box in

Step-by-step explanation:

8 0

3 years ago

After the lunch rush, Ellen had used up all the small containers and half the large containers. Which of these expressions can b

Sergeu [11.5K]

Answer:

$Remaining\ containers=\frac{y}{2}\\\\Where\ y\ is\ the\ number\ of\ large\ containers.$

Step-by-step explanation:

Let number of small containers $=x$

Let number of large containers $=y$

Total containers $=x+y$

<em>Used Containers:</em>

Small containers used $=x\ \ \ (as\ all\ small\ containers\ used)$

Larger containers used $=\frac{y}{2}\ \ \ \ \ \ (half\ of\ the\ large\ containers\ used)$

Total containers used $=x+\frac{y}{2}$

Remaining containers=total containers-used containers

$Remaining\ containers=(x+y)-(x+\frac{y}{2})=\frac{y}{2}\\\\\\\\If\ small\ containers(x)=30\\\\Large\ containers(y)=20\\\\total\ containers(x+y)=30\\\\Remaining\ containers=30-10-\frac{1}{2}\times 20$

6 0

3 years ago

Find the surface area of a cylinder with a radius of 5 cm and a height of 30 cm. Use 3.14

Juli2301 [7.4K]

Whole formula 2*pi*radius^2 + 2*pi*radius*height:

157+942= 1099

SA= 1099

3 0

3 years ago

The Office of Student Services at UNC would like to estimate the proportion of UNC's 28,500 students who are foreign students. I

ANEK [815]

Answer:

For the sampling distribution of the sample proportion for a sample of size 50, the mean is 0.061 and the standard deviation is 0.034.

Step-by-step explanation:

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean $\mu$ and standard deviation $\sigma$ , the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$ .

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

For a proportion p in a sample of size n, the sampling distribution of the sample proportion will be approximately normal with mean $\mu = p$ and standard deviation $s = \sqrt{\frac{p(1-p)}{n}}$