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2 years ago

I have no clue what I’m doing please help! I’m struggling in Graphing Radical Functions!

Mathematics

1 answer:

Dmitriy789 [7]2 years ago

3 0

Answer:

I'm pretty sure you are correct, but it is reflected over the x-axis.

Step-by-step explanation:

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Is it possible for a triangle to have sides with the given lengths: 20 m, 22 m, and 24 m? A. Yes; the sum of each pair is greate

snow_lady [41]

The correct answer is A) yes, the sum of each pair is greater than the third.

The triangle inequality theorem states that the sum of any two sides of a triangle is longer than the third side. This is true for these numbers, so this does form a triangle.

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2 years ago

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Carlo buys $14.40 worth of grapefruit.each grapefruit costs $0.80.

Veronika [31]

He will have bought 18

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2 years ago

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Please help!! I really need this fast! ;(

Nikitich [7]

Answer:

-19

Step-by-step explanation:

if you know subtraction you will know this just subtract 15 from 34 easy can you mark me brainly est. thx

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2 years ago

Erin deposited $400 in a saving account earning 3% yearly compound interest. write an equation to represent how much money (y) E

bekas [8.4K]

$~~~~~~ \textit{Compound Interest Earned Amount \underline{in 15 years}} \\\\ A=P\left(1+\frac{r}{n}\right)^{nt} \quad \begin{cases} A=\textit{accumulated amount}\\ P=\textit{original amount deposited}\dotfill &\$400\\ r=rate\to 3\%\to \frac{3}{100}\dotfill &0.03\\ n= \begin{array}{llll} \textit{times it compounds per year}\\ \textit{yeary, thus once} \end{array}\dotfill &1\\ t=years\dotfill &15 \end{cases}$

$A=400\left(1+\frac{0.03}{1}\right)^{1\cdot 15}\implies A=400(1.03)^{15}\implies A\approx 623.19 \\\\[-0.35em] ~\dotfill\\\\ ~\hfill y=400(1.03)^x~\hfill$

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1 year ago

Three machines A, B, and C produce 55%, 25% and 20% respectively. The percentages of defective output are 3.5%, 4.5% and 5.5%. I

pishuonlain [190]

We are given the following information concerning the three production machines;

$\begin{gathered} Of\text{ the total production;} \\ A=55\text{ \%}=0.55 \\ B=25\text{ \%}=0.25 \\ C=20\text{ \%}=0.20 \end{gathered}$

Also, we are given the percentage of defective output as follows;

$\begin{gathered} A=3.5\text{ \%}=0.035 \\ B=4.5\text{ \%}=0.045 \\ C=5.5\text{ \%}=0.055 \end{gathered}$

Therefore, if an item is selected randomly, the probability that the item is defective would be;

$\begin{gathered} P\lbrack defective\rbrack=(0.55\times0.035)+(0.25\times0.045)+(0.20\times0.055) \\ P\lbrack\text{defective\rbrack}=0.01925+0.01125+0.011 \\ P\lbrack\text{defective\rbrack}=0.0415 \end{gathered}$

ANSWER:

The probability that the item is defective would be 0.0415

6 0

8 months ago