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3 years ago

Some plz help me out. whoever gets it right gets brainliest

Mathematics

1 answer:

tester [92]3 years ago

4 0

Step-by-step explanation:

take care of yourself I hope the help

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You plan to put anti-theft devices on each garment in your store. You have 1,320 garments, and it takes an employee 20 seconds t

makvit [3.9K]

Answer:

3 hours 40 minutes.

Step-by-step explanation:

We are told that we have 1320 garments and it takes an employee 20 seconds to tag a garment.

1 employee tags one garment in 20 seconds.

So garments tagged by 1 employees in 1 minute will be $\frac{60}{20} =3$ garments.

2 employees will tag 3*2=6 garments in 1 minute.

To find the total time it will take two employees to tag all the garments we will divide total number of garments by 6.

$\text{Time taken by two employees to tag 1320 garments}=\frac{1320}{6}$

$\text{Time taken by two employees to tag 1320 garments}=220$

Therefore, it will take 220 minutes or 3 hours and 40 minutes for 2 employees to tag 1320 garments.

6 0

3 years ago

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Can someone please help me??

aivan3 [116]

Answer:

-16 is the answer after evaluating

Step-by-step explanation:

${b }^{2} - 7b - 6 \\ 5 { }^{2} - 7 \times 5 -6 \\ 25 - 35 - 6 \\ 25 - 41 \\ - 16$

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2 years ago

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Please answer this correctly

lozanna [386]

Answer:

no it to low

Step-by-step explanation:

the answer was 4998

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3 years ago

Average precipitation for the first 7 months of the year, the average precipitation in toledo, ohio, is 19.32 inches. if the ave

Colt1911 [192]

Part A:

The probability that a normally distributed data with a mean, μ and standard deviation, σ is greater than a given value, a is given by:

$P(x\ \textgreater \ a)=1-P(x\ \textless \ a)=1-P\left(z\ \textless \ \frac{a-\mu}{\sigma}\right)$

Given that the average precipitation in Toledo, Ohio for the past 7 months is 19.32 inches with a standard deviation of 2.44 inches, the probability that <span>a randomly selected year will have precipitation greater than 18 inches for the first 7 months is given by:

$P(x\ \textgreater \ 18)=1-P(x\ \textless \ 18) \\ \\ =1-P\left(z\ \textless \ \frac{18-19.32}{2.44}\right) \\ \\ =1-P(z\ \textless \ -0.5410) \\ \\ =1-0.29426=\bold{0.7057}$

Part B:

</span>The probability that an n randomly selected samples of a normally distributed data with a mean, μ and standard deviation, σ is greater than a given value, a is given by:

$P(x\ \textgreater \ a)=1-P(x\ \textless \ a)=1-P\left(z\ \textless \ \frac{a-\mu}{\frac{\sigma}{\sqrt{n}}}\right)$

Given that the average precipitation in Toledo, Ohio for the past 7 months is 19.32 inches with a standard deviation of 2.44 inches, the probability that <span>5 randomly selected years will have precipitation greater than 18 inches for the first 7 months is given by:

</span> $P(x\ \textgreater \ 18)=1-P(x\ \textless \ 18) \\ \\ =1-P\left(z\ \textless \ \frac{18-19.32}{\frac{2.44}{\sqrt{5}}}\right) \\ \\ =1-P(z\ \textless \ -1.210) \\ \\ =1-0.1132=\bold{0.8868}$

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3 years ago

Solve x^2 = 196 and c^3 = 216.

Vika [28.1K]

The answer is: B. x=+_14; c=6

Hope this helps, happy holidays!! :p

8 0

2 years ago

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