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Maksim231197 [3]
3 years ago
7

Write a word problem where something starts at 8:25am and ends at 1:43pm​

Mathematics
1 answer:
damaskus [11]3 years ago
5 0

Larry needs to be at work at 8:25 and he gets to leave at 1:43. How long is he ag work?

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damaskus [11]

This is what I got RR || SS

3 0
3 years ago
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Can you help me to solve this<br>​
RUDIKE [14]

Answer:

Is the question even complete ?

Step-by-step explanation:

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1 year ago
What's the answer of {2^3+[4*(10-6)]}/3
Evgesh-ka [11]
Do 2^3 first. It is isolated on one side of a set of the + sign.
{8 + [4*(10 - 6)]} / 3 Next do the innermost 
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{8 + 16} / 3
24/3 Divide by 3
The answer is 8

8 <<<< answer.
4 0
3 years ago
Two competing headache remedies claim to give fast-acting relief. An experiment was performed to compare the mean lengths of tim
DaniilM [7]

Answer:

t = 0.875

Step-by-step explanation:

Given

<u>Brand A</u>            <u>Brand B</u>

n_ 1= 12               n_2 = 12

\bar x_1 = 21.8            \bar x_2 = 18.9

\sigma_1 = 8.7              \sigma_2 = 7.5

Required

Determine the test statistic (t)

This is calculated as:

t = \frac{\bar x_1 - \bar x_2}{s\sqrt{\frac{1}{n_1} +  \frac{1}{n_2}}}

Calculate s using:

s = \sqrt{\frac{(n_1-1)*\sigma_1^2+(n_2-1)*\sigma_2^2}{n_1+n_2-2}}

The equation becomes:

s = \sqrt{\frac{(12-1)*8.7^2+(12-1)*7.5^2}{12+12-2}}

s = \sqrt{\frac{1451.34}{22}}

s = \sqrt{65.97}

s = 8.12

So:

t = \frac{\bar x_1 - \bar x_2}{s\sqrt{\frac{1}{n_1} +  \frac{1}{n_2}}}

t = \frac{21.8 - 18.9}{8.12 * \sqrt{\frac{1}{12} + \frac{1}{12}}}

t = \frac{21.8 - 18.9}{8.12 * \sqrt{\frac{1}{6}}}

t = \frac{21.8 - 18.9}{8.12 * 0.408}}

t = \frac{2.9}{3.31296}}

t = 0.875

3 0
3 years ago
obtain the equation of two points passing of lines 4x-3y-1=0 and 2x-3y+3=0 and equally inclined to the axes​
klasskru [66]

Answer:By 8(4)P.5, the slope of the line equally inclined to axes is tan(±45  

o

)=±1. Hence by P+λQ=0,

5λ+3

2λ+4

​

=±1

⇒λ=  

3

1

​

,−1. Putting for λ, the two lines are

x−y=0,x+y−2=0.

Was this answer helpful

8 0
2 years ago
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