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3 years ago

Help!! it’s a bit blurry but i rlly need help fast

Mathematics

1 answer:

galina1969 [7]3 years ago

3 0

Answer: B

Step-by-step explanation:

It’s the same as the equation given but in half so you would get the same answer.

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a distribution for the last difit of phone numbers sampled from a phone book for the phone number 968-9667 the last difit 7 woul

lisov135 [29]

Answer:

The answer is "Option D".

Step-by-step explanation:

Please find the attached file of the complete question:

In this question, the last choice "Histogram IV" is correct because it would be roughly evenly distributed.

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2 years ago

Find the solution(s) for x in the equation below.

Kobotan [32]

Answer:

D) x=5, x=-5.

Step-by-step explanation:

x^2-25=0

factor

(x+5)(x-5)=0

x+5=0, x-5=0,

x=0-5=-5,

x=0+5=5.

5 0

2 years ago

PLSSS HELP THIS IS HARD ANYONE

Paraphin [41]

Answer:

I think the answer is 3 option

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2 years ago

Read 2 more answers

The standard deviation of the tap water data is 1.086. The standard deviation of the salt water data is 1.107. Which explains wh

Levart [38]

It's A. Nearly Symmetrical.

7 0

3 years ago

Read 2 more answers

At 2 PM, a thermometer reading 80°F is taken outside where the air temperature is 20°F. At 2:03 PM, the temperature reading yiel

Alexeev081 [22]

Use Law of Cooling:
$T(t) = (T_0 - T_A)e^{-kt} +T_A$
T0 = initial temperature, TA = ambient or final temperature

First solve for k using given info, T(3) = 42
$42 = (80-20)e^{-3k} +20 \\ \\ e^{-3k} = \frac{22}{60}=\frac{11}{30} \\ \\ k = -\frac{1}{3} ln (\frac{11}{30})$

Substituting k back into cooling equation gives:
$T(t) = 60(\frac{11}{30})^{t/3} + 20$

At some time "t", it is brought back inside at temperature "x".
We know that temperature goes back up to 71 at 2:10 so the time it is inside is 10-t, where t is time that it had been outside.
The new cooling equation for when its back inside is:
$T(t) = (x-80)(\frac{11}{30})^{t/3} + 80 \\ \\ 71 = (x-80)(\frac{11}{30})^{\frac{10-t}{3}} + 80$
Solve for x:
$x = -9(\frac{11}{30})^{\frac{t-10}{3}} + 80$
Sub back into original cooling equation, x = T(t)
$-9(\frac{11}{30})^{\frac{t-10}{3}} + 80 = 60 (\frac{11}{30})^{t/3} +20$
Solve for t:
$60 (\frac{11}{30})^{t/3} +9(\frac{11}{30})^{\frac{-10}{3}}(\frac{11}{30})^{t/3} = 60 \\ \\ (\frac{11}{30})^{t/3} = \frac{60}{60+9(\frac{11}{30})^{\frac{-10}{3}}} \\ \\ \\ t = 3(\frac{ln(\frac{60}{60+9(\frac{11}{30})^{\frac{-10}{3}}})}{ln (\frac{11}{30})}) \\ \\ \\ t = 4.959$

This means the exact time it was brought indoors was about 2.5 seconds before 2:05 PM

8 0

3 years ago