How long does it take the snail to crawl 73 inches?

Terrance Scott deposited a check for $299.45 and a check for $229.52. He received $42.00 in cash. What was his total deposit?

True [87]

Dude its literally adding come on now lol 570.97

8 0

3 years ago

What is 87 24 over 25 as a decimal

erastova [34]

8724/25 = 348.96
When you say "over" that means that it has to be divided that short subtraction sign that separates the numerator and the denominator and also means divide so I believe that if you divide you can find the decimal.

7 0

3 years ago

What’s the answer to this math problem?plz answer

scZoUnD [109]

Answer:

B

Step-by-step explanation:

Graph B represents a linear function.

Notice that graph D also shows a line, but it is not the graph of a function, since it doesn't pass the vertical line test for functions.

Graph A shows a quadratic function, and graph C shows likely an exponential function. It is graph B the one that is linear, and a function as well.

3 0

4 years ago

Read 2 more answers

A machine is set up to cut metal strips of varying lengths and widths based on the time (t) in minutes. The change in length is

Tcecarenko [31]

If your choices are the following:
a. a(t) = t^4 + 2t
b. a(t) = t^4 + 2t + 3t^(5/2) 
c. a(t) = t^4 - 3t^(5/2) + 2t` 
d. a(t) = t^4 + 2t - 2t^(1/2) + sqrt(t)

The answer is letter c.a(t) = t^4 - 3t^(5/2) + 2t`
Solution:
Area= length times width
Then to get the time, a=a(t), l=l(t), and w=w(t)
So, if Area= l times width
a(t)= l(t) • w(t)
a(t)=[t^2-squrt(t)] • [t^2-2t^1/2]
a(t) = [t^2 - t^(1/2)] • [t^2 - 2t^(1/2)]
=t^4 - 3t^(5/2) + 2t

7 0

3 years ago

Read 2 more answers

Find the conditional probability, in a single roll of two fair 6 sided dive, that the sum is greater than 6, given that neither

arsen [322]

The conditional probability, in a single roll of two fair 6 sided dice, that the sum is greater than 6, given that neither die is a two : 17/25

We know that the conditional probability is given by,

P(B | A) = probability of occurrence of event B, given that event A has

occurred

= P(A ∩ B) / P(A)

Here, P(A ∩ B) means the probability of happening two events A and B at the same time.

We also know that if P (B | A ) = P(B) i.e., P(A ∩ B) = P(A) × P(B) the two events A and B are independent of each other.

For this question, let the dice D1 and D2 are rolled once.

Let the numbers displayed on the dice be d1 and d2 respectively.

The dice D1 and D2 are independent.

We need to find the conditional probability that the sum is greater than 6, given that neither die is a two.

Let S represents the sum of the numbers displayed on the dice.

S = d1 + d2

The sum is even, if d1 = d2 is odd OR if d1 = d2 is even

P(d1 = even) = 3/6

=1/2

P(d2=even) = 1/2

P(d1 = odd) = 1/2

P(d2 = odd) = 1/2

So, P(S = even) = [P(d1=even) × P(d2 = even)] + [P(d1= odd) × P(d2=odd)]

= [1/2 × 1/2] + [1/2 × 1/2]

= 1/2

So, we can say that, the sum is either even or odd which are equally likely and hence its probability is 1/2.

First we find the probability for the sum is greater than 6 i.e., P(S > 6)

The possible combination of d1 and d2 for the sum greater than 6 would be,

{(1,6), (2,5), (2, 6), (3, 4), (3, 5), (3, 6), (4,3), (4,4), (4,5), (4,6), (5,2), (5,3), (5,4), (5,5), (5,6), (6,1), (6,2), (6,3), (6,4), (6,5), (6,6)}

⇒ n(S > 6) = 21

The number of all possible outcomes = 36

So, P(S > 6) = 21/36

= 7/12

Now we find the probability that neither die is a two

⇒ P(neither die is a two) = [P(1) ∪ P(3 ≤ d1 ≤ 6)] AND [P(1) ∪ P(3 ≤ d1 ≤ 6)]

⇒ P(neither die is a two) = 5/6 × 5/6

⇒ P(neither die is a two) = 25/36

Now, we find the probability that the sum S > 6 AND neither die is a two.

The possible combination for the sum S > 6 AND neither die is a two would be,

{(1,6), (3, 4), (3, 5), (3, 6), (4,3), (4,4), (4,5), (4,6), (5,3), (5,4), (5,5), (5,6), (6,1), (6,3), (6,4), (6,5), (6,6)}

⇒ n(S > 6 AND neither die is a two) = 17

So, P(S > 6 AND neither die is a two) = 17/36

Now we find the conditional probability P(S > 6 | neither die is a two)

⇒ P(S > 6 | neither die is a two) = P(S > 6 AND neither die is a two) ÷

(neither die is a two)

⇒ P(S > 6 | neither die is a two) = (17/36) / (25/36)

⇒ P(S > 6 | neither die is a two) = 17/25

Therefore, the conditional probability, in a single roll of two fair 6 sided dice, that the sum is greater than 6, given that neither die is a two : 17/25

Learn more about the probability here:

brainly.com/question/3679442

#SPJ4

4 0

1 year ago