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4 years ago

Which table represents a direct variation function?

Mathematics

2 answers:

Luda [366]4 years ago

5 0

Direct variation is y=kx where k is a constant

the fiest way to see if it is direct or not, is if x increases, then y increases as well,
then we see if y=kx is valid, basically if we have a constant of variation

the first one x increase and y increase
see if same constant
y=kx
-4.5=-3k
1.5=k
so

see next one
-1 and 3
-3=-1(k)
-3=-1(1.5)
-3=-1.5
false
not it

2nd is increase and y decrease, so not direct variation

3rd is x is same but y increase so nope

4th is x increase and y increase, now test the constant
-7.5=-3k
2.5=k

-1 and -2.5
-2.5=-1k
-2.5=-1(2.5)
-2.5=-2.5
true

answer is last option

Liula [17]4 years ago

3 0

Answer:

D on ed

Step-by-step explanation:

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Finn changes his mind and, from now on, decides to take the normal route to work everyday. On any given day, the time (in minute

Margaret [11]

Answer:

The 33rd percentile of the time it takes Finn to get to work on any given day is 31.04 minutes.

There is a 61.92% probability that Finn took more than 40.5 minutes to get to work on the first day or more than 38.5 minutes to get to work on the second day.

Step-by-step explanation:

This can be solved by the the z-score formula:

On a normaly distributed set with mean $\mu$ and standard deviation $\sigma$ , the z-score of a value X is given by:

$Z = \frac{X - \mu}{\sigma}$

Each z-score value has an equivalent p-value, that represents the percentile that the value X is:

The problem states that:

Mean = 35, so $\mu = 35$

Variance = 81. The standard deviation is the square root of the variance, so $\sigma = \sqrt{81} = 9$ .

Find the 33rd percentile of the time it takes Finn to get to work on any given day. Do not include any units in your answer.

Looking at the z-score table, $z = -0.44$ has a pvalue of 0.333. So what is the value of X when $z = -0.44$ .

$Z = \frac{X - \mu}{\sigma}$

$-0.44 = \frac{X - 35}{9}$

$X - 35 = -3.96$

$X = 31.04$

The 33rd percentile of the time it takes Finn to get to work on any given day is 31.04 minutes.

Over the next 2 days, find the probability that Finn took more than 40.5 minutes to get to work on the first day or more than 38.5 minutes to get to work on the second day.

$P = P_{1} + P_{2}$

$P_{1}$ is the probability that Finn took more than 40.5 minutes to get to work on the first day. The first step to solve this problem is finding the z-value of $X = 40.5$ .

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{40.5 - 35}{9}$

$Z = 0.61$

$Z = 0.61$ has a pvalue of 0.7291. This means that the probability that it took LESS than 40.5 minutes for Finn to get to work is 72.91%. The probability that it took more than 40.5 minutes if $P_{1} = 100% - 72.91% = 27.09% = 0.2709$

$P_{2}$ is the probability that Finn took more than 38.5 minutes to get to work on the second day. Sine the probabilities are independent, we can solve it the same way we did for the first day, we find the z-score of

$X = 38.5$

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{38.5 - 35}{9}$

$Z = 0.39$

$Z = 0.39$ has a pvalue of 0.6517. This means that the probability that it took LESS than 38.5 minutes for Finn to get to work is 65.17%. The probability that it took more than 38 minutes if $P_{1} = 100% - 65.17% = 34.83% = 0.3483$

So:

$P = P_{1} + P_{2} = 0.2709 + 0.3483 = 0.6192$

There is a 61.92% probability that Finn took more than 40.5 minutes to get to work on the first day or more than 38.5 minutes to get to work on the second day.

7 0

4 years ago

Linn plans to wallpaper the 4 walls in her bedroom. The measurements are 9’ by 11’ by 8’. How many single rolls of wallpaper sho

Ann [662]

Answer: There are 4 rolls she should buy.

Step-by-step explanation:

Since we have given that

Length of wall = 9 feet

Width of wall = 11 feet

Height of wall = 8 feet

As we know the formula for "Area of four walls ":

$Area=2(l+b)h\\\\Area=2(9+11)\times 8\\\\Area=2\times 20\times 8\\\\Area=320\ ft^2$

Area of each single roll covers = 72 square feet

So, Number of single rolls of wallpaper is given by

$\frac{\text{ Area of 4 walls}}{\text{ Area of single roll}}\\\\=\frac{320}{72}\\\\=4.44\\\\=4$

Hence, there are 4 rolls she should buy.

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4 years ago

5v-2v-v simply this equation

Dmitriy789 [7]

Answer:

Step-by-step explanation:

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3 years ago

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The half-life of cesium-137 is 30 years. Suppose we have a

VMariaS [17]

(a) If $y(t)$ is the mass (in mg) remaining after $t$ years, then $y(t) = y(0) e^{kt}=100e^{kt}.$

$y(30) = 100 e^{30k} = \frac{1}{2}(100) \implies e^{30k} = \frac{1}{2} \implies k = -(\ln 2) /30 \implies \\ \\
y(t) = 100e^{-(\ln 2)t/30} = 100 \cdot 2^{-t/30}$

(b) $y(100) = 100 \cdot 2^{-100/30} \approx \text{9.92 mg}$

(c)
$100 e^{- (\ln 2)t/30} = 1\ \implies\ -(\ln 2) t / 30 = \ln \frac{1}{100}\ \implies\ \\ \\
t = -30 \frac{\ln 0.01}{\ln 2} \approx \text{199.3 years}$

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3 years ago

Find the accumulated value after 12 years of deposits of $360 made at the beginning of every 3 months and earning interest of 7.

Effectus [21]

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