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Olenka [21]
3 years ago
14

Please help fast. Due at 9:35 today . Hurry

Mathematics
1 answer:
OLEGan [10]3 years ago
7 0
Shush shush shush shusb
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John ordered five large size pizzas for a party. Five kids each ate 1/10 slice, and twelve adults each at 3/10 slices.
astraxan [27]

Answer:

a) 4 1/10 slices eaten

b) cannot be solved because we don't know how many slices in each pizza

Step-by-step explanation:

5*1/10=5/10

12*3/10=36/10

36/10+5/10=41/10

41/10=4 1/10 slices

8 0
3 years ago
7. The graph shows the height, h, in feet, of water
MariettaO [177]

The domain is talking about the "range" of the horizontal axis therefore you will be focusing on the x-intercepts.

The answer will be All non-negative real numbers less than or equal to 18

because the x-intercepts lies at 0 and 18. The answer makes sense because the furthest you can go is 18 ft and the closest you could go is 0 ft. The "all non-negative real numbers" puts a restriction on the least distance it could travel so that means that it stops at 0 ft because if you go any further, you will end up in the negatives and it clearly states "non-negative".

4 0
2 years ago
Determine whether the following conjecture is true or false. Give a counterexample if the answer is false.
brilliants [131]
This conjecture is false, A and C don't have to be complementary angles.

Let Angle A be 100 and Angle B be 80 degrees. They are supplementary (sum of 180).
Let Angle C be 100 and Angle B is still 80 degrees. They are supplementary.

However, Angle A and Angle C are not complementary (sum of 90).
4 0
3 years ago
A box contains 3 coins. One coin has 2 heads and the other two are fair. A coin is chosen at random from the box and flipped. If
Blababa [14]

Answer: Our required probability is \dfrac{1}{2}

Step-by-step explanation:

Since we have given that

Number of coins = 3

Number of coin has 2 heads = 1

Number of fair coins = 2

Probability of getting one of the coin among 3 = \dfrac{1}{3}

So, Probability of getting head from fair coin = \dfrac{1}{2}

Probability of getting head from baised coin = 1

Using "Bayes theorem" we will find the probability that it is the two headed coin is given by

\dfrac{\dfrac{1}{3}\times 1}{\dfrac{1}{3}\times \dfrac{1}{2}+\dfrac{1}{3}\times \dfrac{1}{2}+\dfrac{1}{3}\times 1}\\\\=\dfrac{\dfrac{1}{3}}{\dfrac{1}{6}+\dfrac{1}{6}+\dfrac{1}{3}}\\\\=\dfrac{\dfrac{1}{3}}{\dfrac{2}{3}}\\\\=\dfrac{1}{2}

Hence, our required probability is \dfrac{1}{2}

No, the answer is not \dfrac{1}{3}

5 0
3 years ago
Find tan A and tan B.
Alex

Step-by-step explanation:

\tan(a)  =  \frac{26}{10}  = 2.6 \\  \tan(b)  =  \frac{10}{26}

8 0
3 years ago
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