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3 years ago

Find the y- intercept of the following equation y = 6x + 10 Please help quick!

Mathematics

1 answer:

n200080 [17]3 years ago

8 0

The slope is 6 and the y intercept is 10. Mark brainliest would be aporeciated

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Put the numbers in order from smallest to largest.

AnnZ [28]

In order of problems/numbers - 4, 3, 5, 6, 1, 2

3 0

3 years ago

Please help me with this math question!!!!!

astraxan [27]

Answer:

Step-by-step explanation:

7x+30+9x+42=104

16x+72=104

16x=104-72

16x=32

x=2

SO:

<BOC=9x+42=

9(2)+42=

18+42=

7 0

3 years ago

Inferential statistics allow you to decide whether a difference between the experimental and the control group is due to:_______

den301095 [7]

Answer:

Option (A)

Step-by-step explanation:

There are 2 main branches of statistics. They are:

1. Descriptive Statistics

2. Inferential Statistics

Descriptive statistics describes the characteristics of observed subjects or items while Inferential statistics makes inferences, based on given or derived data.

Inferential Statistics allow you to decide whether a difference between the experimental group and control group is due to <u>manipulation or chance.</u>

<u />

The Experimental group is the group that an effect is tested on while the Control group is the group that is left untested or uninfluenced. Inferential statistics allow you to decide whether a difference between these 2 groups is due to

- manipulation or interference by any force (which may be the experimenter/researcher)

- probability which is chance.

5 0

3 years ago

Ok ok this i need help

lakkis [162]

Answer:

Step-by-step explanation:

18-4^2=2

7 0

3 years ago

Read 2 more answers

Suppose Kaitlin places $6500 in an account that pays 12% interest compounded each year.

Leya [2.2K]

$\bf ~~~~~~ \textit{Compound Interest Earned Amount \underline{for 1 year}} \\\\ A=P\left(1+\frac{r}{n}\right)^{nt} \quad \begin{cases} A=\textit{accumulated amount}\\ P=\textit{original amount deposited}\dotfill &\$6500\\ r=rate\to 12\%\to \frac{12}{100}\dotfill &0.12\\ n= \begin{array}{llll} \textit{times it compounds per year}\\ \textit{annually, thus once} \end{array}\dotfill &1\\ t=years\dotfill &1 \end{cases}$

$\bf A=6500\left(1+\frac{0.12}{1}\right)^{1\cdot 1}\implies A=6500(1.12)\implies A=7280 \\\\[-0.35em] ~\dotfill$

$\bf ~~~~~~ \textit{Compound Interest Earned Amount \underline{for 2 years}} \\\\ A=P\left(1+\frac{r}{n}\right)^{nt} \quad \begin{cases} A=\textit{accumulated amount}\\ P=\textit{original amount deposited}\dotfill &\$6500\\ r=rate\to 12\%\to \frac{12}{100}\dotfill &0.12\\ n= \begin{array}{llll} \textit{times it compounds per year}\\ \textit{annually, thus once} \end{array}\dotfill &1\\ t=years\dotfill &2 \end{cases}$

$\bf A=6500\left(1+\frac{0.12}{1}\right)^{1\cdot 2}\implies A=6500(1.12)^2\implies A=8153.6$

6 0

3 years ago