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Rasek [7]
2 years ago
10

Galen wrote the statement "If the sum of the digits in a number is divisible by 3, the original number is divisible by 3." Which

statements are true regarding his work? Check all that apply.
A: The hypothesis of the statement is "If the sum of the digits in a number is divisible by 3."

B: An equivalent statement is "If the sum of the digits in a number is divisible by 3, then the original number is divisible by 3."

C: The statement is not a conditional statement because it does not include both an "if" and a "then" clause.

D:The statement can be proven true or false.

E:The conclusion of the statement is written before the hypothesis.
Mathematics
1 answer:
lord [1]2 years ago
6 0

the answer is B. good luck

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The SAT mathematics scores in the state of Florida are approximately normally distributed with a mean of 500 and a standard devi
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Since the range of the scores given was between 300 and 700 (which is 2 standard deviations below and above the mean), the probability that a randomly selected student's math score - as based on the empirical rule of statistics - is 95%. In decimal form, it is .95. 
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1 year ago
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The aquarium at Sea Critters Depot contains 140 fish. Eighty of these fish are green swordtails (44 female and 36 male) and 60 a
maxonik [38]

Given that:

Total number of fish = 140

Fish are green swordtails female = 44

Fish are green swordtails male = 36

Fish are orange swordtails female = 36

Fish are orange swordtails male = 24

Solution:

A. We have to find the probability that the selected fish is a green swordtail.

\text{P(green swordtail)}=\dfrac{\text{Total green swordtail fish}}{\text{Total fish}}

\text{P(green swordtail)}=\dfrac{80}{140}

\text{P(green swordtail)}=\dfrac{4}{7}

Therefore, the probability that the selected fish is a green swordtail is \dfrac{4}{7}.

B.  We have to find the probability that the selected fish is male.

\text{P(Male fish)}=\dfrac{\text{Total male fish}}{\text{Total fish}}

\text{P(Male fish)}=\dfrac{36+24}{140}

\text{P(Male fish)}=\dfrac{60}{140}

\text{P(Male fish)}=\dfrac{3}{7}

Therefore, the probability that the selected fish is a male, is \dfrac{3}{7}.

C. We have to find the probability that the selected fish is a male green swordtail.

\text{P(Male green swordtail)}=\dfrac{\text{Total male green swordtail fish}}{\text{Total fish}}

\text{P(Male green swordtail)}=\dfrac{36}{140}

\text{P(Male green swordtail)}=\dfrac{9}{35}

Therefore, probability that the selected fish is a male green swordtail is \dfrac{9}{35}.

D.

We have to find the probability that the selected fish is either a male or a green swordtail.

\text{P(Male or green swordtail)}=\dfrac{\text{Total male or green swordtail fish}}{\text{Total fish}}

\text{P(Male or green swordtail)}=\dfrac{44+36+24}{140}

\text{P(Male or green swordtail)}=\dfrac{96}{140}

\text{P(Male or green swordtail)}=\dfrac{24}{35}

Therefore, the probability the selected fish is either a male or a green swordtail is \dfrac{24}{35}.

4 0
2 years ago
At smith’s Mountain Lake Boat Rentals, it cost $25 per hour to rent a pontoon boat, plus a one-time
sergey [27]

Answer:

Jones family paid a total of $139

Step-by-step explanation:

Smith's Mountain Lake Boat provides Rental services that can be expressed as follows;

Total cost=cost per hour×number of hours rented+one time cleaning deal

For Benael's family;

Benael family total cost=Cost per hour×number of hours rented+one-time cleaning deal

where;

Benael family total cost=$226.50

Cost per hour=$25

Number of hours rented=11 am-6:30 pm=7 hours 30 minutes=7.5 hours

One time cleaning deal=x

Replacing;

226.50=(25×7.5)+x

187.5+x=226.50

x=226.50-187.5

x=39

One time cleaning deal=$39

For Jones family;

Jones family total cost=Cost per hour×number of hours rented+one-time cleaning deal

where;

Cost per hour=$25

Number of hours rented=9 am-1 pm=4 hours

One time cleaning deal=$39

Replacing;

Jones family total cost=(25×4)+39

Jones family total cost=$139

Jones family paid a total of $139

4 0
2 years ago
In a study of the progeny of rabbits, Fibonacci (ca. 1170-ca. 1240) encountered the sequence now bearing his name. The sequence
expeople1 [14]

The question in part c is not clear, nevertheless, part a and part b would be solved.

Answer:

a. The first twelve terms are:

1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144.

b. The first ten terms are:

1.000, 1.000, 1.500, 1.667, 1.600, 1.625, 1.615, 1.619, 1.618, 1.618.

Step-by-step explanation:

a. Given

an + 2 = an + an + 1

where a1 = 1 and a2 = 1.

a3 = a1 + a2

= 2

a4 = a2 + a3

= 3

a5 = a3 + a4

= 5

a6 = a5 + a4

= 8

a7 = a6 + a5

= 13

a8 = a7 + a6

= 21

a9 = a8 + a7

= 34

a10 = a9 + a8

= 55

a11 = a10 + a9

= 89

a12 = a11 + a10

= 144

The first twelve terms are:

1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144.

(b)

Given

bn = an+1/an

b1 = a2/a1

= 1/1 = 1.000

b2 = a3/a2

= 2/1 = 1.000

b3 = a4/a3

= 3/2 = 1.500

b4 = a5/a4

= 5/3 = 1.667

b5 = a6/a5

= 8/5 = 1.600

b6 = a7/a6

= 13/8 = 1.625

b7 = a8/a7

= 21/13 = 1.615

b8 = a9/a8

= 34/21 = 1.619

b9 = a10/a9

= 55/34 = 1.618

b10 = a11/a10

= 89/55 = 1.618

The first ten terms are:

1.000, 1.000, 1.500, 1.667, 1.600, 1.625, 1.615, 1.619, 1.618, 1.618.

6 0
1 year ago
The probability that Naoya succeeds at any given free-throw is 70%, percent. He was curious how many free-throws he can expect t
Arturiano [62]

Answer:

Step-by-step explanation:

P(greater than 10 successes)≈  

13/20 =0.65

7 0
2 years ago
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