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3 years ago

F(x)=(1.07)^

Mathematics

1 answer:

Shalnov [3]3 years ago

8 0

F(x ) = 1.07^x
Is an exponential growth function of 7%

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Kevin and Randy Muise have a jar containing 70 coins, all of which are either quarters or nickels. The total value of the coins

Nikitich [7]

Answer:

50 nickels, 20 quarters.

Step-by-step explanation:

System of equations (q = # of quarters, n = # of nickels):

q + n = 70, 0.25q + 0.05n = 7.50

the first equation can be changed to q = 70 - n, so we are able to substitute q with 70 - n.

So, it will look like 0.25*70 - 0.25n + 0.05n = 7.50. This can be simplified to 0.2n = 10, which means that n = 50.

Knowing that we can solve q + 50 = 70, which means that q = 20.

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3 years ago

Answer the question attached: (I need this ASAP)

Illusion [34]

Answer:

A≈534.07

Step-by-step explanation:

A=2πrh+2πr^2

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3 years ago

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Which of the following is a mixed number?

Elanso [62]

"a" would be the correct answer

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3 years ago

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Question 1

stiks02 [169]

Answer:

Step-by-step explanation:

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3 years ago

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A student records the number of hours that they have studied each of the last 23 days. They compute a sample mean of 2.3 hours a

natita [175]

Answer:

the standard deviation increases

Step-by-step explanation:

Let x₁ , x₂, . . . , x₂₃ be the actual data observed by the student

The sample means = x₁ + x₂ + . . . , x₂₃ / 23

$= \frac{x_1 +x_2 +...x_2_3}{23}$

= 2.3hr

⇒ $\sum xi =2.3 \times 23 = 52.9hrs$

let x₁ , x₂, . . . , x₂₃ arranged in ascending order

Then x₂₃ was 10 and has been changed to 14

i.e x₂₃ increase to 4

Sample mean $= \frac{x_1 +x_2 +...x_2_3}{23}$

$\frac{52.9hrs + 4}{23} \\\\= \frac{56.9}{23} \\\\= 2.47$

therefore, the new sample mean is 2.47

2) For the old data set

the median is $x_1_2(th)$ values

$[\frac{n +1}{2} ]^t^h value$

when we use the new data set only x₂₃ is changed to 14

i.e the rest all observation remain unchanged

Hence, sample median = $[{x_1_2]^t^h value$ remain unchange

sample median = 2.5hrs

The Standard deviation of old data set is calculated

$=\sqrt{\frac{1}{n-1} \sum (xi - \bar x_{old})^2 } \\\\=\sqrt{\frac{1}{22}\sum ( xi - 2.3)^2 }---(1)$

The new sample standard sample deviation is calculated as

$= \sqrt{\frac{1}{n-1} \sum (xi-2.47)^2} ---(2)$

Now, when we compare (1) and (2) the square distance between each observation xi and old mean is less than the squared distance between each observation xi and the new mean.

Since,

(xi - 2.3)² ∑ (xi - 2.47)²

Therefore , the standard deviation increases

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4 years ago