jake needs 180 in of wallpaper border. he has three and 1 half yards. how many more yards does he need?

There is an animal farm where chickens and cows live. All together, there are 84 heads and 210 legs. How many chickens and cows

Ira Lisetskai [31]

1 Chicken have = 2 legs

1 Cow have = 4 legs

No of cows = x

No of chickens = y

Total no of legs = 4x + 2y = 210.............(1)

Total No of heads = x + y = 84................(2)

x = 84 - y....( from (2) )

Substituting the above equation in (1)

4 ( 84 - y) + 2y = 210

336 - 4y + 2y = 210

336 - 2y = 210

-2y = -126

= y = 63

No of cows = 84-63 = 21

∴ Total no of cows = 21

Total no of chickens = 63

5 0

2 years ago

Hhelp plssssssssssssssssssssssssssss

Ipatiy [6.2K]

Answer:

This is false

Step-by-step explanation:

7 0

2 years ago

Six cards numbered from 1 to 6 are placed in an empty bowl. First one card is drawn and then put back into the bowl; then a seco

kari74 [83]

Answer:

C. $\frac{1}{18}$

Step-by-step explanation:

Given: Six cards numbered from $1$ to $6$ are placed in an empty bowl. First one card is drawn and then put back into the bowl then a second card is drawn.

To Find: If the cards are drawn at random and if the sum of the numbers on the cards is $8$ , what is the probability that one of the two cards drawn is numbered $5$ .

Solution:

Sample space for sum of cards when two cards are drawn at random is $\{(1,1),(1,2),(1,3)......(6,6)\}$

total number of possible cases $=36$

Sample space when sum of cards is $8$ is $\{(3,5),(5,3),(6,2),(2,6),(4,4)\}$

Total number of possible cases $=5$

Sample space when one of the cards is $5$ is $\{(5,3),(3,5)\}$

Total number of possible cases $=2$

Let A be the event that sum of cards is $8$

$p(\text{A})$ $=\frac{\text{total cases when sum of cards is 8}}{\text{all possible cases}}$

$p(\text{A})=\frac{5}{36}$

Let B be the event when one of the two cards is $5$

probability than one of two cards is $5$ when sum of cards is $8$

$p(\frac{\text{B}}{\text{A}})=\frac{\text{total case when one of the number is 5}}{\text{total case when sum is 8}}$

$p(\frac{\text{B}}{\text{A}})=\frac{2}{5}$

Now,

probability that sum of cards $8$ is and one of cards is $5$

$p(\text{A and B}=p(\text{A})\times p(\frac{\text{B}}{\text{A}})$

$p(\text{A and B})=\frac{5}{36}\times\frac{2}{5}$

$p(\text{A and B})=\frac{1}{18}$

if sum of cards is $8$ then probability that one of the cards is $8$ is $\frac{1}{18}$ , option C is correct.

3 0

2 years ago

275 miles on 14 gallons pls hurry

vaieri [72.5K]

Answer:

1 gallon per 19.64mi

Step-by-step explanation:

275/14 =19.6428571429

8 0

2 years ago

Read 2 more answers

Given the function f(x) = x2 and k = 3, which of the following represents the graph becoming more narrow? A. f(x)+k

Neko [114]

The only type of transformation that can make a graph more narrow/wide is a scaling transformation. A scaling transformation involves a multiple factor. The answer to your question is B. I hope that this is the answer that you were looking for and it has helped you.

4 0

2 years ago

Read 2 more answers