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3 years ago

A tree was 9 feet tall. One year later, the tree was 16 feet tall. Write an equation and find how many feet f the tree grew

Mathematics

2 answers:

ArbitrLikvidat [17]3 years ago

8 0

Answer:

9 + f = 16

Step-by-step explanation:

So just say 9 + f = 16 since the tree grew to 16, and the f is the amount.

Bingel [31]3 years ago

5 0

Answer:

16-9=7

Step-by-step explanation:

the tree grew 7 ft

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Suppose it is known that the distribution of purchase amounts by customers entering a popular retail store is approximately norm

iragen [17]

Answer:

a. 0.691

b. 0.382

c. 0.933

d. $88.490

e. $58.168

f. 5th percentile: $42.103

95th percentile: $107.897

Step-by-step explanation:

We have, for the purchase amounts by customers, a normal distribution with mean $75 and standard deviation of $20.

a. This can be calculated using the z-score:

$z=\dfrac{X-\mu}{\sigma}=\dfrac{85-75}{20}=\dfrac{10}{20}=0.5\\\\\\P(X$

The probability that a randomly selected customer spends less than $85 at this store is 0.691.

b. We have to calculate the z-scores for both values:

$z_1=\dfrac{X_1-\mu}{\sigma}=\dfrac{65-75}{20}=\dfrac{-10}{20}=-0.5\\\\\\z_2=\dfrac{X_2-\mu}{\sigma}=\dfrac{85-75}{20}=\dfrac{10}{20}=0.5\\\\\\\\P(65$

The probability that a randomly selected customer spends between $65 and $85 at this store is 0.382.

c. We recalculate the z-score for X=45.

$z=\dfrac{X-\mu}{\sigma}=\dfrac{45-75}{20}=\dfrac{-30}{20}=-1.5\\\\\\P(X>45)=P(z>-1.5)=0.933$

The probability that a randomly selected customer spends more than $45 at this store is 0.933.

d. In this case, first we have to calculate the z-score that satisfies P(z<z*)=0.75, and then calculate the X* that corresponds to that z-score z*.

Looking in a standard normal distribution table, we have that:

$P(z$

Then, we can calculate X as:

$X^*=\mu+z^*\cdot\sigma=75+0.67449\cdot 20=75+13.4898=88.490$

75% of the customers will not spend more than $88.49.

e. In this case, first we have to calculate the z-score that satisfies P(z>z*)=0.8, and then calculate the X* that corresponds to that z-score z*.

Looking in a standard normal distribution table, we have that:

$P(z>-0.84162)=0.80$

Then, we can calculate X as:

$X^*=\mu+z^*\cdot\sigma=75+(-0.84162)\cdot 20=75-16.8324=58.168$

80% of the customers will spend more than $58.17.

f. We have to calculate the two points that are equidistant from the mean such that 90% of all customer purchases are between these values.

In terms of the z-score, we can express this as:

$P(|z|$

The value for z* is ±1.64485.

We can now calculate the values for X as:

$X_1=\mu+z_1\cdot\sigma=75+(-1.64485)\cdot 20=75-32.897=42.103\\\\\\X_2=\mu+z_2\cdot\sigma=75+1.64485\cdot 20=75+32.897=107.897$

5th percentile: $42.103

95th percentile: $107.897

5 0

4 years ago

The real numbers $x$ and $y$ are such that \begin{align*} x + y &= 4, \\ x^2 + y^2 &= 22, \\ x^4 &= y^4 - 176 \sqrt{

tamaranim1 [39]

You get everything you need from factoring the last expression:

$x^4-y^4=-176\sqrt7$

The left side is a difference of squares, and we get another difference of squares upon factoring. We end up with

$x^4-y^4=(x^2-y^2)(x^2+y^2)=(x-y)(x+y)(x^2+y^2)$

Plug in everything you know and solve for $x-y$ :

$-176\sqrt7=(x-y)\cdot4\cdot22\implies x-y=\boxed{-2\sqrt7}$

4 0

3 years ago

Read 2 more answers

The manager wants to advertise that anybody who isn't served within a certain number of minutes gets a free hamburger. But she d

aksik [14]

Answer:

The number of minutes advertisement should use is found.

x ≅ 12 mins

Step-by-step explanation:

(MISSING PART OF THE QUESTION: AVERAGE WAITING TIME = 2.5 MINUTES)

For such problems, we can use probability density function, in which probability is found out by taking integral of a function across an interval.

Probability Density Function is given by:

$f(t)=\left \{ {{0 ,\-t$

Consider the second function:

$f(t)=\frac{e^{-t/\mu}}{\mu}\\$

Where Average waiting time = μ = 2.5

The function f(t) becomes

$f(t)=0.4e^{-0.4t}$

The manager wants to give free hamburgers to only 1% of her costumers, which means that probability of a costumer getting a free hamburger is 0.01

The probability that a costumer has to wait for more than x minutes is:

$\int\limits^\infty_x {f(t)} \, dt= \int\limits^\infty_x {}0.4e^{-0.4t}dt$