Josie saw a sign advertising 6 ears of sweet corn for $1.50. how much would 15 ears of corn cost?

101 E Victoria Ct, Greenville, NC 27858 my address from mr beast

RideAnS [48]

Answer:

uh what is this address for or am i not supposed to be in this

Step-by-step explanation:

sorry if im getting in your buisness

3 0

2 years ago

Read 2 more answers

If U = the set of odd natural numbers and B = {13, 15, 17, 19, 21, 23, . . .}, find B′.

Yuki888 [10]

Answer:

{1,3,5,7,9,11}

Step-by-step explanation:

B' stands for the complement of B.

B' means set elements in the universe that aren't in B.

Here is the universe:

U={1,3,5,7,9,11,13,15,17,19,21,23,25,27,29,31,33,...}

(The dots mean it continues with that pattern forever and ever.)

And the elements we taking out:

B={13,15,17,19,21,23,...}

So we taking out the odd numbers from U that are odd and greater than or equal to 13. (The dots mean it continues with that pattern forever and ever.)

So B' is the set of natural odd numbers that are less than 13:

{1,3,5,7,9,11}

8 0

3 years ago

Read 2 more answers

Solve for y in -8×+4y=24

s344n2d4d5 [400]

Factor out the common term 4

-4(2x - y) = 24

Divide both sides by -4

2x-y= - 24/4

Simplify24/4 to 6

2x -y = -6

Subtract 2x from both sides

-y = -6 - 2x

Multiply both sides by -1

y = 6 + 2x

5 0

3 years ago

Suppose on a certain planet that a rare substance known as Raritanium can be found in some of the rocks. A raritanium-detector i

aleksandrvk [35]

Answer:

0.967 = 96.7% probability the rock sample actually contains raritanium

Step-by-step explanation:

Conditional Probability

We use the conditional probability formula to solve this question. It is

$P(B|A) = \frac{P(A \cap B)}{P(A)}$

In which

P(B|A) is the probability of event B happening, given that A happened.

$P(A \cap B)$ is the probability of both A and B happening.

P(A) is the probability of A happening.

In this question:

Event A: Positive reading

Event B: Contains raritanium

Probability of a positive reading:

98% of 13%(positive when there is raritanium).

0.5% of 100-13 = 87%(false positive, positive when there is no raritanium). So

$P(A) = 0.98*0.13 + 0.005*0.87 = 0.13175$

Positive when there is raritanium:

98% of 13%

$P(A) = 0.98*0.13 = 0.1274$

What is the probability the rock sample actually contains raritanium?

$P(B|A) = \frac{P(A \cap B)}{P(A)} = \frac{0.1274}{0.13175} = 0.967$

0.967 = 96.7% probability the rock sample actually contains raritanium

7 0

2 years ago

Please help I want summer school done. Which graph corresponds to the given function?

Alexus [3.1K]

Answer:

The first option in the first picture

Step-by-step explanation:

Good luck with summer school bud :)

3 0

2 years ago