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Luden [163]
3 years ago
9

The low temperature in Leroy’s town one day was -4 Fahrenheit . The difference between the high temperature and the low temperat

ure that day was 6 Fahrenheit The equation h(-4)=6 can be used to find h, the high temperature in the town that day in degrees Fahrenheit. What was the high temperature in the town that day, in degrees Fahrenheit?
Mathematics
2 answers:
Nitella [24]3 years ago
6 0

Answer:

6-4=2 The high temperature was 2 degrees Fahrenheit

Step-by-step explanation:

Im Smart

EastWind [94]3 years ago
3 0
-4 + 6 = 2, so the high that day was 2 degrees. Since we're finding the high, and -4 is the low, it would not make sense for us to subtract 6 from -4, which would result in an even lower number. Thus, we add to find out the high.

Hope this helps! Have a great day :) 
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Step-by-step explanation:

10 gallons of gas is $29.50

10 gallons of gas = $29.50

10 gallons of gas / 10 = $29.50 / 10

1 gallon of gas = $2.95

Per x gallons of gas,

1 gallon of gas * x = $2.95 * x

x gallons of gas = $2.95 * x

7 0
3 years ago
Please help giving BRANLIEST to the best answer THIS SHOULD BE EASY
solniwko [45]

Answer:

the size difference of figure two to figure one gets smaller

Step-by-step explanation:

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3 years ago
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What is the term of the highest degree in the expression 3x2y-5xy7+8x4y5-6xy
skelet666 [1.2K]

Answer:

seems hard

Step-by-step explanation:

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4 0
3 years ago
Simplify:<br><br> -5x - 2(3x - 7) + 4(3x - 5) - (6x -9)
eduard

Answer:

7x+3

Step-by-step explanation:

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7 0
2 years ago
The physical plant at the main campus of a large state university receives daily requests to replace fluorescent lightbulbs. The
Nikolay [14]

Answer:

P(62< X< 72)= P(X

Step-by-step explanation:

The empirical rule, also known as three-sigma rule or 68-95-99.7 rule, "is a statistical rule which states that for a normal distribution, almost all data falls within three standard deviations (denoted by σ) of the mean (denoted by µ)".

Let X the random variable who represent the courtship time (minutes).

From the problem we have the mean and the standard deviation for the random variable X. E(X)=62, Sd(X)=5

So we can assume \mu=62 , \sigma=5

On this case in order to check if the random variable X follows a normal distribution we can use the empirical rule that states the following:

• The probability of obtain values within one deviation from the mean is 0.68

• The probability of obtain values within two deviation's from the mean is 0.95

• The probability of obtain values within three deviation's from the mean is 0.997

So we need values such that

P(X    

P(X>\mu +\sigma)=P(X >67)=0.16  

P(X    

P(X>\mu +2*\sigma)=P(X>72)=0.025

P(X

P(X>\mu +3*\sigma)=P(X>77)=0.0015

For this case we want to find this probability:

P(62 < X< 72)

And we can find this probability on this way:

P(62< X< 72)= P(X

Since P(X>72) =0.025 by the complement rule we have that:

P(X

And P(X because for this case 62 is the mean.

So then we have this:

P(62< X< 72)= P(X

7 0
3 years ago
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