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ExtremeBDS [4]
3 years ago
12

Please solve this! Need it to finish my class. Giving Brainliest to whoever gets it right!

%5Csqrt%7B18%7D%20-%20%5Cfrac%7B3%7D%7B8%7D" id="TexFormula1" title="4*\sqrt{18} - \frac{3}{8}" alt="4*\sqrt{18} - \frac{3}{8}" align="absmiddle" class="latex-formula">
Mathematics
1 answer:
Afina-wow [57]3 years ago
7 0

Answer:

square root of 282, and in decimal form its 16.79, rounded

Step-by-step explanation:

hi

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Brut [27]

Answer:

Probability that Hannah only has to buy 3 or less boxes before getting a prize is 0.784

Step-by-step explanation:

Given, 40% of cereal boxes contain a prize

⇒probability of getting a prize on opening a box, P(A)=0.4

where A is the event of getting a prize on opening a cereal box

and probability of not getting a prize on opening a box, P(A')=1-P(A)=0.6

where A' is the event of not getting a prize on opening a cereal box

This problem needs to be divided into 3 situation:

  • Case 1, Where Hannah gets prize when she buys the first box:

Let K be the event of Hannah winning the prize on buying the first box.

⇒P(K)=P(A)=0.4

  • Case 2, Where Hannah gets prize when she buys the second box:

I<u>n this event Hannah should not get the prize in first box but should get the prize on buying the second box</u>

Let L be the event of Hannah winning the prize on buying the second box

So, P(L)=P(A')·P(A)

           =(0.6)·(0.4)

           =0.24

  • Case 3,Where Hannah gets prize when she buys the third box:

<u>In this event Hannah should not get the prize in first and second box but should get the prize on buying the third box</u>

Let L be the event of Hannah winning the prize on buying the third box

So, P(L)=P(A')·P(A')·P(A)

           =(0.6)·(0.6)·(0.4)

           =0.144

Let N be the event of Hannah winning the prize on buying 3 or less boxes before getting a prize

⇒N=K∪L∪M

Now, Required probability is P(N)=P(K∪L∪M)=P(K)+P(L)+P(M) [As events K,L and M are independent and disjoint events]

⇒P(N)=0.4+0.24+0.144

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Tom had 8 cookies, and he ate one and a half cookies. how many cookies were left?
Alex777 [14]
Tom had 6 and a half cookies
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3 years ago
Probability of a standard six side dice landing on a four in the first three rolls?
Illusion [34]

Answer:

91/216

Step-by-step explanation:

The probability of getting a 4 in the first three rolls is 1 minus the probability of not getting a 4 on any of the rolls.

P(at least one 4) = 1 − P(no 4s)

P(at least one 4) = 1 − (5/6)³

P(at least one 4) = 91/216

Alternatively, you can calculate it this way.

The probability of getting a 4 on the first roll is 1/6.

The probability of getting a 4 on the second roll is (5/6) (1/6) = 5/36.

The probability of getting a 4 on the third roll is (5/6) (5/6) (1/6) = 25/216.

The probability of any of the three events is 1/6 + 5/36 + 25/216 = 91/216.

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Answer:

stop

Step-by-step explanation:

stop using this app, the chances of you actually getting a helpful response or even one at all are so slim, just pay an actual tutor at this point

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