a sailfish can go as fast as 68mph. in 1 min. can a sailfish swim as far as 1 mile? explain the answer.

Help meh emergency. No links please give a right answer, need helppppp

andre [41]

Answer:

70.91 x 10 = 709.1

70.91 x 100 = 7,091

70.91 x 1000 = 70,910

8 0

3 years ago

Read 2 more answers

What percent of 56 is 8?

34kurt

Answer:

Step-by-step explanation:

56/8=7

4 0

3 years ago

Plug in 3 for the equation y=-x^2+12x-20

notka56 [123]

Step-by-step explanation:

y=(-x^2)+12x-20

=(-(3)^2)+12(3)-20

=(-9)+36-20

=16-9=7

y=7

I hope it helped you

8 0

2 years ago

The gross weekly sales at a certain restaurant are a normal random variable with mean $2200 and standard deviation $230.What is

Lesechka [4]

Answer:

a) The total gross sales over the next 2 weeks exceeds $5000 is 0.0321.

b) The weekly sales exceed $2000 in at least 2 of the next 3 weeks is 0.9033.

Step-by-step explanation:

Given : The gross weekly sales at a certain restaurant are a normal random variable with mean $2200 and standard deviation $230.

To find : What is the probability that

(a) the total gross sales over the next 2 weeks exceeds $5000;

(b) weekly sales exceed $2000 in at least 2 of the next 3 weeks? What independence assumptions have you made?

Solution :

Let $X_1$ and $X_2$ denote the sales during week 1 and 2 respectively.

a) Let $X=X_1+X_2$

Assuming that $X_1$ and $X_2$ follows same distribution with same mean and deviation.

$E(X)=E(X_1+X_2)=E(X_1)+E(X_2)$

$E(X)=2\mu = 2(220)=4400$

$\sigma_X=\sqrt{var(X_1+X_2)}$

$\sigma_X=\sqrt{2\sigma^2}$

$\sigma_X=\sqrt{2}\sigma$

$\sigma_X=230\sqrt{2}$

So, $X\sim N(4400,230\sqrt{2})$

$P(X>5000)=1-P(X\leq5000)$

$P(X>5000)=1-P(Z\leq\frac{5000-4400}{230\sqrt{2}})$

$P(X>5000)=1-P(Z\leq1.844)$

$P(X>5000)=1-0.967$

$P(X>5000)=0.0321$

The total gross sales over the next 2 weeks exceeds $5000 is 0.0321.

b) The probability that sales exceed teh 2000 and amount in at least 2 and 3 next week.

We use binomial distribution with n=3.

$P(X>2000)=1-P(X\leq2000)$

$P(X>2000)=1-P(Z\leq\frac{2000-2200}{230})$

$P(X>2000)=1-P(Z\leq-0.87)$

$P(X>2000)=1-0.1922$

$P(X>2000)=0.808$

Let Y be the number of weeks in which sales exceed 2000.

Now, $P(Y\geq 2)$

So, $P(Y\geq 2)=P(Y=2)+P(Y=3)$

$P(Y\geq 2)=^3C_2(0.8077)^2\cdot (1-0.8077)+^3C_3(0.8077)^3$

$P(Y\geq 2)=0.37635+0.52692$

$P(Y\geq 2)=0.90327$

The weekly sales exceed $2000 in at least 2 of the next 3 weeks is 0.9033.

3 0

3 years ago

Nate tosses a ball up a hill for his dog to chase. The path of the ball is modeled by the function y = –14x2 + 335x, where x is

tankabanditka [31]

Answer:

The horizontal distance the ball travels before it hits the ground is 22. $\overline{857142}$ feet

Step-by-step explanation:

The given parameter are;

The function modelling the path of the ball tossed by Nate y = -14·x² + 335·x

x = The horizontal distance the ball travels from Nate in feet

y = The height of the ball in feet

The line equation modelling the hill is y = 15·x

The point where the ball hits the ground is given by the point the graph of the equation for the path of the ball and the path of the model of the line of the hill meet as follows;

Ball path is y = -14·x² + 335·x

Hill path is y = 15·x

The point both paths meet and the ball hits the ground is -14·x² + 335·x = 15·x

Which gives;

-14·x² + 335·x - 15·x = 0

-14·x² + 320·x = 0

320·x - 14·x² = 0

x × (320 - 14·x) = 0

x = 0, or x = 320/14 = 22 6/7 = 22. $\overline{857142}$ feet

Therefore;

The horizontal distance the ball travels before it hits the ground = x = 22. $\overline{857142}$ feet.

3 0

3 years ago