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Evgesh-ka [11]
2 years ago
11

Using spreadsheets to Organize Data (PLEASE HELP)

Mathematics
1 answer:
katrin [286]2 years ago
6 0
I can give you the first half!
1. D2 - hours worked on monday, 3hrs
2. E2 - money earned on monday, $15
3. D3 - hours worked on tuesday, 3 hrs
4. E3 - money earned on tuesday, $15
5. D4 - hours worked on thursday, 4 hrs
6. E4 - money earned on thursday, $20
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Find an equation of a line with slope -3/4 and that passes through P(-4,6) in standard, form.
SashulF [63]

Answer:

y = -   \frac{3}{4}x  + 3

Step-by-step explanation:

the standard form of any straight line is:

y = m × x + c

where: y is the y value of the point

x is the x value of the point

m is the gradient/slope

SOLUTION

y = mx + c \\ 6 =  -  \frac{3}{4} ( - 4) + c \\ c = 3

3 0
2 years ago
Show all relevant steps.
Usimov [2.4K]
The answer is A hope this helps!
7 0
3 years ago
Say that a supplier claims they are 99% confident that their products will be in the interval of 50.02 to 50.38. You take sample
nydimaria [60]

Answer:

The supplier becomes less accurate than they otherwise would have tried to claim. A further explanation is below.

Step-by-step explanation:

According to the provider, this same width of that same confidence interval would be as follows:

= 50.38 - 50.02

= 0.36

Depending on the input observed, the width including its confidence interval would be as follows:

= 50.39 - 50.01

= 0.38

As even the width of that interval again for survey asserted > the width including its confidence interval according to the provider's statement, we could conclude that such is the appropriate reaction.

3 0
3 years ago
How do I do this? I’m so confused
Shtirlitz [24]

I say left because you have a straight line, i goes through 0,0 and every time it goes over 1 and up by 3.

5 0
4 years ago
How many 10-digit ternary strings are there that contain exactly two 0s, three 1s, and five 2s?
Svet_ta [14]

There are \dbinom{10}2 ways of picking 2 of the 10 available positions for a 0. 8 positions remain.

There are \dbinom83 ways of picking 3 of the 8 available positions for a 1. 5 positions remain, but we're filling all of them with 2s, and there's \dbinom55=1 way of doing that.

So we have

\dbinom{10}2\dbinom83\dbinom55=\dfrac{10!}{2!(10-2)!}\dfrac{8!}{3!(8-3)!}\dfrac{5!}{5!(5-5)!}=2520

The last expression has a more compact form in terms of the so-called multinomial coefficient,

\dbinom{10}{2,3,5}=\dfrac{10!}{2!3!5!}=2520

5 0
3 years ago
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