Ask question

Ask question

All categories

2 years ago

15

If you drive 8 miles south then make a left turn and drive 6 miles east how far are you in a straight line from your starting po

int

1 answer:

olganol [36]2 years ago

4 0

Answer:

10 miles

Step-by-step explanation:

d=√8²+6²=√100= 10 miles

You might be interested in

Given the function ƒ(x ) = 3x + 1, evaluate ƒ(a + 1).

Setler79 [48]

Answer:

$\Huge \boxed{3a+4}$

Step-by-step explanation:

The function is given.

$f(x)=3x+1$

To find : $f(a+1)$

The input for the function f(x) is (a + 1).

Replace x with (a + 1).

$f(a+1)=3(a+1)+1$

Expand brackets.

$f(a+1)=3a+3+1$

Simplifying.

$f(a+1)=3a+4$

3 0

3 years ago

Cevabını yazarmısınız

Anit [1.1K]

Answer:

it is side ways

Step-by-step explanation:

6 0

3 years ago

Cards are drawn, one at a time, from a standard deck; each card is replaced before the next one is drawn. Let X be the number of

steposvetlana [31]

Cards are drawn, one at a time, from a standard deck; each card is replaced before the next one is drawn. Let X be the number of draws necessary to get an ace. Find E(X) is given in the following way

Step-by-step explanation:

From a standard deck of cards, one card is drawn. What is the probability that the card is black and a jack? P(Black and Jack) P(Black) = 26/52 or ½ , P(Jack) is 4/52 or 1/13 so P(Black and Jack) = ½ * 1/13 = 1/26

A standard deck of cards is shuffled and one card is drawn. Find the probability that the card is a queen or an ace.

P(Q or A) = P(Q) = 4/52 or 1/13 + P(A) = 4/52 or 1/13 = 1/13 + 1/13 = 2/13

WITHOUT REPLACEMENT: If you draw two cards from the deck without replacement, what is the probability that they will both be aces?

P(AA) = (4/52)(3/51) = 1/221.

WITHOUT REPLACEMENT: What is the probability that the second card will be an ace if the first card is a king?

P(A|K) = 4/51 since there are four aces in the deck but only 51 cards left after the king has been removed.

WITH REPLACEMENT: Find the probability of drawing three queens in a row, with replacement. We pick a card, write down what it is, then put it back in the deck and draw again. To find the P(QQQ), we find the

probability of drawing the first queen which is 4/52.

The probability of drawing the second queen is also 4/52 and the third is 4/52.

We multiply these three individual probabilities together to get P(QQQ) =

P(Q)P(Q)P(Q) = (4/52)(4/52)(4/52) = .00004 which is very small but not impossible.

Probability of getting a royal flush = P(10 and Jack and Queen and King and Ace of the same suit)

5 0

3 years ago

A telephone booth that is 8 ft tall casts a shadow that is 4ft long. Find the height of a lawn ornanment that casts a 2 ft shado

sukhopar [10]

8 ft /4 ft of shadow=x / 2 ft of shadow
x=(8 ft * 2 ft of shadow) / 4 ft of shadow=4 ft

height of a lawn ornament=4 ft

5 0

3 years ago

Let X denote the temperature (degree C) and let Y denote thetime in minutes that it takes for the diesel engine on anautomobile

BlackZzzverrR [31]

Answer:

Step-by-step explanation:

Given $f_{XY} (x,y) = c(4x + 2y +1) ; 0 < x < 40\,and\, 0 < y$

a)

we know that $\int\limits^\infty_{-\infty}\int\limits^\infty_{-\infty} {f(x,y)} \, dxdy=1$

therefore $\int\limits^{40}_{-0}\int\limits^2_{0} {c(4x+2y+1)} \, dxdy=1$

on integrating we get

c=(1/6640)

b)

$P(X>20, Y>=1)=\int\limits^{40}_{20}\int\limits^2_{1} {\frca{1}{6640}(4x+2y+1)} \, dxdy$

on doing the integration we get

=0.37349

c)

marginal density of X is

$f(x)=\int\limits^2_{0} {\frca{1}{6640}(4x+2y+1)} \, dy$

on doing integration we get

f(x)=(4x+3)/3320 ; 0<x<40

marginal density of Y is

$f(y)=\int\limits^{40}_{0} {\frca{1}{6640}(4x+2y+1)} \, dx$

on doing integration we get

$f(y)=\frac{(y+40.5)}{83}$

d)

$P(01)=\int\limits^{40}_{0}\int\limits^2_{1} {\frca{1}{6640}(4x+2y+1)} \, dxdy$

solve the above integration we get the answer

e)

$P(X>20, 0$

solve the above integration we get the answer

f)

Two variables are said to be independent if there jointprobability density function is equal to the product of theirmarginal density functions.

we know f(x,y)

In the (c) bit we got f(x) and f(y)

f(x,y)cramster-equation-2006112927536330036287f(x).f(y)

therefore X and Y are not independent

4 0

3 years ago