Rewrite 9 5/2 in radical form.

ki77a [65]

3x + 4y is simplified

8 0

3 years ago

A boat on a river travels downstream between two points, 90 mi apart, in 1 h. The return trip against the current takes 2 1 2 h.

docker41 [41]

Answer:

A)63miles per hour.

B)27 miles per hour

Step-by-step explanation:

HERE IS THE COMPLETE QUESTION

boat on a river travels downstream between two points, 90 mi apart, in 1 h. The return trip against the current takes 2 1 2 h. What is the boat's speed (in still water)??b) How fast does the current in the river flow?

Let the speed of boat in still water = V(boat)

speed of current=V(current)

To calculate speed of boat downstream, we add speed of boat in still water and speed of current. This can be expressed as

[V(boat) +V(current)]

It was stated that it takes 1hour for the

boat to travels between two points of 90 mi apart downstream.

To calculate speed of boat against current, we will substact speed of current from speed of boat in still water. This can be expressed as

[V(boat) - V(current)]

and it was stated that it takes 2 1/2 for return trip against the Current

But we know but Speed= distance/time

Then if we input the stated values we have

V(boat) + V(current)]= 90/1 ---------eqn(1)

V(boat) - V(current) = 90/2.5----------eqn(2)

Adding the equations we have

V(boat) + V(current) + [V(boat) - V(current)]= 90/2.5 + 90/1

V(boat) + V(current) + V(boat) - V(current)]=90+36

2V(boat)= 126

V(boat)=63miles per hour.

Hence, Therefore, the speed of boat in still water is 63 miles per hour.

?b) How fast does the current in the river flow?

the speed of the current in the river, we can be calculated if we input V(boat)=63miles per hour. Into eqn(1)

V(boat) + V(current)]= 90/1

63+V(current)=90

V(current)= 27 miles per hour

Hence,Therefore, the speed of current is 27 miles per hour.

7 0

3 years ago

Help<br> Pls :/.........

Sveta_85 [38]

Answer:

Step-by-step explanation:

m<4= 109*

m<3= 71*

m<1= 71*

3 0

3 years ago

Duck #1 lays eggs whose weights are normally distributed with a mean of 70gramsand a standard deviation of 6 grams.Duck #2 lays

Genrish500 [490]

Answer:

26.11% probability that Duck #2’s egg weighs more than Duck #1’s egg

Step-by-step explanation:

When the distribution is normal, we use the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

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

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

If an egg is randomly chosen from each duck, what is the probability that Duck #2’s egg weighs more than Duck #1’s egg?

Duck #2 egg will weigh more if the subtraction of duck's 2 egg by duck's 1 egg is larger than 0.

When we subtract normal distributions, the mean is the subtraction of the means. So

$\mu = 65 - 70 = -5$

The standard deviation is the square root of the sum of the variances. So

$\sigma = \sqrt{6^2+5^2} = \sqrt{61} = 7.81$

Now, we have to find 1 subtracted by the pvalue of Z when X = 0. So

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

$Z = \frac{0 - (-5)}{7.81}$

$Z = 0.64$

$Z = 0.64$ has a pvalue of 0.7389

1 - 0.7389 = 0.2611

26.11% probability that Duck #2’s egg weighs more than Duck #1’s egg

4 0

3 years ago

A website manager has noticed that during the evening hours, about 5 people per minute check out from their shopping cart and m

Over [174]

Answer:

a) Poisson distribution

b) 99.33% probability that in any one minute at least one purchase is made

c) 0.05% probability that seven people make a purchase in the next four minutes

Step-by-step explanation:

In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by the following formula:

$P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}$

In which

x is the number of sucesses

e = 2.71828 is the Euler number

$\mu$ is the mean in the given time interval.

5 people per minute check out from their shopping cart and make an online purchase.

This means that $\mu = 5$

a) What model might you suggest to model the number of purchases per minute? 

The only information that we have is the mean number of an event(purchases) in a time interval. Each event is also independent fro each other. So you should suggest the Poisson distribution to model the number of purchases per minute.

b) What is the probability that in any one minute at least one purchase is made? 

Either no purchases are made, or at least one is. The sum of the probabilities of these events is 1. So

$P(X = 0) + P(X \geq 1) = 1$