Ask question

Ask question

All categories

3 years ago

12

Pls help give u a free brainlist

1 answer:

Brilliant_brown [7]3 years ago

3 0

Answer:

correct answer is J

Step-by-step explanation:

You might be interested in

What is the y intercept

Diano4ka-milaya [45]

Y intercept is (0, -4/5)
Set x as 0 and solve
Ignore the dumb kid who said y=8

3 0

3 years ago

The sum of two numbers is 60. The greater number is 6 more than the smaller number. Which equation can be used to solve for the

WARRIOR [948]

Answer:

1) x + (x + 6) = 60

Step-by-step explanation:

<u>Let the numbers be x and y</u>

x + y = 60
y = x + 6

<u>Then, substituting y:</u>

x + (x + 6) = 60

Correct option is 1)

7 0

3 years ago

Solve the equation using the zero-product property. (2x − 8)(7x + 5) = 0 x = –2 or x = 7 x = 4 or x = x = 4 or x = x = –4 or x =

aliya0001 [1]

$(2x-8)(7x+5)=0\iff2x-8=0\ \vee\ 7x+5=0\\\\2x-8=0\qquad\text{add 8 to both sides}\\2x=8\qquad\text{divide both sides by 2}\\\boxed{x=4}\\\\7x+5=0\qquad\text{subtract 5 from both sides}\\7x=-5\qquad\text{divide both sides by 7}\\\boxed{x=-\dfrac{5}{7}}$

5 0

3 years ago

Janae' is buying food for a party. She is buying a cheese tray for $4.50 and a few pounds of fruit for

bogdanovich [222]

Answer:

0.67

Step-by-step explanation:

5 0

3 years ago

Two distinct number cubes, one red and one blue, are rolled together. Each number cube has sides numbered 1 through 6.

cestrela7 [59]

Answer:

$P(E_1 or E_2) = \frac{7}{12}$

Step-by-step explanation:

Given

Two cubes of side 1 - 6

Required

Probability that the outcome of the roll is an odd sum or a sum that is a multiple of 5

First, the sample space needs to be listed;

Let $C_r$ represent the Red cube

$C_b$ represent the Blue cube

S represent the sample space

$C_r = (1,2,3,4,5,6)\\C_b = (1,2,3,4,5,6)\\S = (2,3,4,5,6,7,3,4,5,6,7,8,4,5,6,7,8,9,5,6,7,8,9,10,6,7,8,9,10,11,7,8,9,10,11,12)$ S is gotten by getting the sum of $C_r$ and $C_b$

$n(S) = 36$

<em>Calculating the Probability</em>

Let $E_1$ represent the event that an outcome is an odd sum

$E_1 = (3,5,7,3,5,7,5,7,9,5,7,9,7,9,11,7,9,11)$

$n(E_1) = 18$

$P(E_1) = \frac{n(E_1)}{n(S)}$

$P(E_1) = \frac{18}{36}$

Let $E_2$ represent the event that an outcome is a multiple of 5

$E_2 = (5,5,5,5,10,10,10)$

$n(E_2) = 7$

$P(E_2) = \frac{n(E_2)}{n(S)}$

$P(E_2) = \frac{7}{36}$

Let $E_3$ represent the event that an outcome is an odd sum and a multiple of 5

$E_3 = E_1 and E_2$

$E_3 = (5,5,5,5)$

$n(E_3) = 4$

$P(E_3) = \frac{n(E_3)}{n(S)}$

$P(E_3) = \frac{4}{36}$

Calculating $P(E_1 or E_2)$

$P(E_1 or E_2) = P(E_1) + P(E_2) - P(E_1 and E_2)$

$P(E_1 or E_2) = P(E_1) + P(E_2) - P(E_3)$

$P(E_1 or E_2) = \frac{18}{36} + \frac{7}{36} - \frac{4}{36}$

$P(E_1 or E_2) = \frac{18 + 7 - 4}{36}$

$P(E_1 or E_2) = \frac{21}{36}$

$P(E_1 or E_2) = \frac{7}{12}$

Hence, the probability that the outcome of the roll is an odd sum or a sum that is a multiple of 5 is $\frac{7}{12}$

6 0

3 years ago