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Basile [38]
2 years ago
11

Clara and Toby are telemarketers.

Mathematics
1 answer:
Korolek [52]2 years ago
3 0

Answer:

Toby will reach more people in 40 phone calls

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PLEASE HELP!!!!! How many 4-digit numbers divisible by 5, all of the digits of which are even, are there?
saul85 [17]

Answer:

I assume you know Arithmetic Progression .

so, we have to find the first and last 4-digit number divisible by 5

first = 1000 ,  last =  9990

we have a formula,   a_{n} = a + (n-1)d

here, a_{n} is the last 4-digit number divisible by 5.

n is the number of 4-digit even numbers divisible by 5

d is the common difference between the numbers, which is 10 in this case

a is the first 4-digit number divisible by 5

9990 = 1000 + (n-1)*10

899 = n-1

n = 900

Hence, there are 900 4-digit even numbers divisible by 5

7 0
2 years ago
Giving thanks and brainliest for best answer <3
UNO [17]
The answer is D 80-99
4 0
2 years ago
Read 2 more answers
HELP PLEASE HELP HELP
kupik [55]
I think it’s 0but not s
5 0
2 years ago
how much material was used in the manufacture of 24000 celluloid dice,if each dice has an edge 1/4 inches?
Dovator [93]
<span>The volume of each celluloid die is (.25 x .25 x .25) = 0.015625 cubic inch.

To manufacture 24,000 of them, you need to start with <u>at least</u>

(24,000) x (0.015625) = <u>375 cubic inches</u>.

I don't know how celluloid is sold, so you should also keep in mind that
375 cubic inches = about 207.8 fluid ounces, or about 6.5 quarts.

I'm sure a bit more than that was used in the manufacture, since
there's always some wasted, spilled, or trimmed off of the edges. </span>
4 0
3 years ago
A coin is tossed 5 times. Find the probability that all are heads. Find the probability that at most 2 are heads.
fenix001 [56]

Answer:

1/32

15/32

Step-by-step explanation:

For a fair sided coin,

Probability of heads, P(H) = 1/2

Probability of tails P(T)  = 1/2

For a coin tossed 5 times,

P( All heads)

= P(HHHHH),

= P (H) x P(H) x P(H) x P(H) x P(H)

= (1/2) x (1/2) x (1/2) x (1/2) x (1/2)

= 1/32 (Ans)

For part B, it is easier to just list the possible outcomes for

"at most 2 heads" aka "could be 1 head" or "could be 2 heads"

"One Head" Outcomes:

P(HTTTT), P(THTTT) P(TTHTT), P(TTTHT), P(TTTTH)

"2 Heads" Outcomes:

P(HHTTT), P(HTHTT), P(HTTHT), P(HTTTH), P(THHTT), P(THTHT), P(THTTH), P(TTHHT), P(TTHTH), P(TTTHH)

If we count all the possible outcomes, we get 15 possible outcomes representing "at most 2 heads)

we know that each outcome has a probability of 1/32

hence 15 outcomes for "at most 2 heads" have a probability of

(1/32) x 15  = 15/32

8 0
3 years ago
Read 2 more answers
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