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erik [133]
2 years ago
3

the mean temperature for the past 10 days was 22 degrees c. If the sum if the temperature for the first nine days was 200, what

was the temperature on day 10? A.) 10c B.) 32c C.) 22c D) 20c​
Mathematics
1 answer:
Mumz [18]2 years ago
6 0

Answer:

D) 20c

Step-by-step explanation:

Mean is the average of a set of numbers.  To find the average, you add up the values of the data and divide by the number of data points.  In this case, the number of data points is 10 days and the average was 22 degrees.  Given that the sum of the first nine days was 200, you can let 'x' be the temperature on the tenth day and solve for the temperature:

\frac{200+x}{10}=22

Multiply 10 on both sides: 200 + x = 220

Subtract 200 on both sides: 200 + x - 200 = 220 - 200

Solve for x: x = 20c

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We have an arithmetic progression:
Nth=an
an=a₁+(n-1)d
a₁ is the first term.
n=number of terms.
d=common difference

10,17,24,31...
a₁=10
d=a₂-a₁=17-10=7

Therefore:
Nth=an
an=a₁+(n-1)d
an=10+(n-1)7
an=10+7n-7
an=7n+3.

Therefore: the formula for the nth is, an=a+(n-1), in this case; an=7n+3,

To check:
a₁=7*1+3=10
a₂=7*2+3=17
a₃=7*3+3=24
a₄=7*4+3=31
a₅=7*5+3=38.......
8 0
2 years ago
f1(x) = ex, f2(x) = e−x, f3(x) = sinh(x) g(x) = c1f1(x) + c2f2(x) + c3f3(x) Solve for c1, c2, and c3 so that g(x) = 0 on the int
eimsori [14]

Answer:

(C1, C2, C3) = (K, K, -2K)

For K in the interval (-∞, ∞)

Step-by-step explanation:

Given

f1(x) = e^x

f2(x) = e^(-x)

f3(x) = sinh(x)

g(x) = 0

We want to solve for C1, C2 and C3, such that

C1f1(x) + C2f2(x) + C3f3(x) = g(x)

That is

C1e^x + C2e^(-x) + C3sinh(x) = 0

The hyperbolic sine of x, sinh(x), can be written in its exponential form as

sinh(x) = (1/2)(e^x + e^(-x))

So, we can rewrite

C1e^x + C2e^(-x) + C3sinh(x) = 0

as

C1e^x + C2e^(-x) + C3(1/2)(e^x + e^(-x)) = 0

So we have

(C1 + (1/2)C3)e^x + (C2 + (1/2)C3)e^(-x) = 0

We know that

e^x ≠ 0, and e^(-x) ≠ 0

So we must have

(C1 + (1/2)C3) = 0...........................(1)

and

(C2 + (1/2)C3) = 0..........................(2)

From (1)

2C1 + C3 = 0

=> C3 = -2C1.................................(3)

From (2)

2C2 + C3 = 0

=> C3 = -2C2................................(4)

Comparing (3) and (4)

2C1 = 2C2

=> C2 = C1

Let C1 = C2 = K

C3 = -2K

(C1, C2, C3) = (K, K, -2K)

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3 0
2 years ago
2. A computer shop charges P20.00 per hour (or a fraction of an hour) for the first
Misha Larkins [42]

Answer:

(a) For 40 minutes the charge is P20.00

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The amount the computer shop charges for the first two hours = P20.00

The amount the computer shop charges for a fraction of an hour = P20.00

The amount the computer shop charges for each extra hour = P10.00

Therefore, the amount the computer shop will charge for each time, is given as follows;

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The total amount charged for the 3 hours = P20.00 + P10.00 = P30.00

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150 minutes =  2 hours 30 minutes

Therefore, the amount charged will be equivalent to the amount charged for 3 hours or P20.00 + P10.00 = P30.00.

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