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Elden [556K]
3 years ago
14

suppose that during one period of extreme cold the average daily temp decreased 1 1/2 Fahrenheit each day how many days did it t

ake for the temp to decrease by 9 degrees Fahrenheit for barrow Alaska
Mathematics
1 answer:
Nina [5.8K]3 years ago
7 0

Answer:

The temperature decreased by 9 °F in <u>6 days.</u>

Step-by-step explanation:

Let the number of days it takes for the temperature to decrease by 9 °F be x.

Given:

Average daily temperature decreases by 1\frac{1}{2} °F.

Decrease of temperature in 1 day = 1\frac{1}{2}= 1.5 °F.

∴ Decrease in temperature in x days is 1.5x.

Now, as per question, decrease in temperature in 'x' days is 9 °F.

Therefore, 1.5x=9

x=\frac{9}{1.5}

x=6

Therefore, the temperature decreased by 9 °F in 6 days.

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Using partial product, are the products 48 and 64
shusha [124]
For 49 It's 4*12 and 2*24 and 1*48 and for 64 it's 8*8 and 2*32
4 0
3 years ago
The amount of time a passenger waits at an airport check-in counter is random variable with mean 10 minutes and standard deviati
Stolb23 [73]

Answer:

(a) less than 10 minutes

= 0.5

(b) between 5 and 10 minutes

= 0.5

Step-by-step explanation:

We solve the above question using z score formula. We given a random number of samples, z score formula :

z-score is z = (x-μ)/ Standard error where

x is the raw score

μ is the population mean

Standard error : σ/√n

σ is the population standard deviation

n = number of samples

(a) less than 10 minutes

x = 10 μ = 10, σ = 2 n = 50

z = 10 - 10/2/√50

z = 0 / 0.2828427125

z = 0

Using the z table to find the probability

P(z ≤ 0) = P(z < 0) = P(x = 10)

= 0.5

Therefore, the probability that the average waiting time waiting in line for this sample is less than 10 minutes = 0.5

(b) between 5 and 10 minutes

i) For 5 minutes

x = 5 μ = 10, σ = 2 n = 50

z = 5 - 10/2/√50

z = -5 / 0.2828427125

= -17.67767

P-value from Z-Table:

P(x<5) = 0

Using the z table to find the probability

P(z ≤ 0) = P(z = -17.67767) = P(x = 5)

= 0

ii) For 10 minutes

x = 10 μ = 10, σ = 2 n = 50

z = 10 - 10/2/√50

z = 0 / 0.2828427125

z = 0

Using the z table to find the probability

P(z ≤ 0) = P(z < 0) = P(x = 10)

= 0.5

Hence, the probability that the average waiting time waiting in line for this sample is between 5 and 10 minutes is

P(x = 10) - P(x = 5)

= 0.5 - 0

= 0.5

3 0
3 years ago
The ratio 5:20 written as a fraction​
iogann1982 [59]

Answer:

<em> </em><em>1</em><em>:</em><em>4</em>

<em>hope</em><em> </em><em>helpful</em><em> </em>:) !!!!

4 0
3 years ago
Read 2 more answers
Using​ 45-in. fabric, Regan needs 2 and one fourth yd for a​ dress, one fourth yd of contrasting fabric for the band at the​ bot
Arte-miy333 [17]

Answer:

7\dfrac14$ yards

Step-by-step explanation:

Regan needs the following:

2\dfrac{1}{4} yards for a​ dress

\dfrac{1}{4}$ yards of contrasting fabric for the band at the bottom$\\4\dfrac{3}{4}$ yards for a coordinating jacket.

Total Length of fabric needed

= 2\dfrac{1}{4}+\dfrac{1}{4}+4\dfrac{3}{4}\\=2+4+\dfrac{1}{4}+\dfrac{3}{4}+\dfrac{1}{4}\\=6+1+\dfrac{1}{4}\\=7\dfrac{1}{4}$ yards

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4 0
3 years ago
Read 2 more answers
It is estimated that the annual cost of driving a certain new car is given by the formula C = 0.37m + 2600 where m represents th
arsen [322]

Answer:

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Step-by-step explanation:

We are given that the annual cost of driving a certain car is given by the formula

C=0.37m+2600

Where m=Represents the number of miles driven per year

C=Cost in dollars

We have to find the range of  miles that Jane  can drive her new car.

Range of budgets=$7040-$7780

Substitute C=$7040 in the given formula

7040=0.37m+2600

7040-2600=0.37m

4440=0.37m

m=\frac{4440}{0.37}

m=12000

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7780=0.37m+2600

7780-2600=0.37m

5180=0.37m

m=\frac{5180}{0.37}=14000

The range of miles that Jane can drive her new car=12000-14000

4 0
3 years ago
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