What name best describes this shape?

Mathematics

1 answer:

EastWind [94]3 years ago

7 0

Answer: Quadrilateral

Explanation:

Any four sided polygon is considered a quadrilateral since "quad" means "four". You can think of a quad bike, which has four wheels, to help remember this math term.

Usually in problems like this, we'll be given information about if certain sides are parallel or the same length. However, we don't this info, so we cannot narrow down this classification anymore.

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Simplify (343) 1/3 a)21 b)19 c)49 d)7

Aliun [14]

I think the answer is D

5 0

3 years ago

What is the slope of a line that passes through (6,4) and (-1, 3)

pshichka [43]

Answer:

C) 1/7

Step-by-step explanation:

so idk what the formula is called but it looks like this:

y2-y1/x2-x1 and if you plug in the numbers then you'll get the answer

3 - 4/-1 - 6 = -1/-7 = 1/7

you can also switch the numbers your answer will always be the same

4 - 3/6 - (-1) = 1/7

so the answer is c

4 0

4 years ago

Can someone help me with this

hichkok12 [17]

Answer:

on mathaway you can ask it to solve using a table of values.

Or if you have a graphing calculator you can hit the 2nd, and then the graph buttons to get a table of values

Step-by-step explanation:

4 0

3 years ago

A local bank needs information concerning the savings account balances of its customers. A random sample of 15 accounts was chec

PSYCHO15rus [73]

Answer:

(513.14, 860.36) is a 98% confidence interval for the true mean

Step-by-step explanation:

We have a small sample size of n = 15, the mean balance was $\bar{x} = 686.75$ with a standard deviation of s = 256.20. The confidence interval is given by $\bar{x}\pm t_{\alpha/2}(\frac{s}{\sqrt{n}})$ where $t_{\alpha/2}$ is the 100 $(\alpha/2)$ th quantile of the t distribution with n-1=15-1=14 degrees of freedom. As we want a $100(1-\alpha)$ % = 98% confidence interval, we have that $\alpha = 0.02$ and the confidence interval is $686.75\pm t_{0.01}(\frac{256.20}{\sqrt{15}})$ where $t_{0.01}$ is the 1st quantile of the t distribution with 14 df, i.e., $t_{0.01} = -2.6245$ . Then, we have $686.75\pm (-2.6245)(\frac{256.20}{\sqrt{15}})$ and the 98% confidence interval is given by (513.1379, 860.3621)