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3 years ago

What is 1.3h + 3.7 when h = 5

Mathematics

2 answers:

MakcuM [25]3 years ago

8 0

Answer:

10.2

Step-by-step explanation:

1.3h + 3.7

1.3 x 5 = 6.5

6.5 + 3.7 = 10.2

Anvisha [2.4K]3 years ago

3 0

Answer:

The answer is 10.20

Hope this helps!! :)

Step-by-step explanation:

1.3 * h(5)= 6.5

6.5 + 3.7= 10.20

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How do solve (3/8+2/3)+1/3 how do you work the problem

Eddi Din [679]

Get rid of parenthesis, this is addition it doesn't matter how you add it will give you the same answer
3/8+2/3+1/3
3/8+3/3
3/8+1
3/8+8/8
11/8→1 3/8→1.375

5 0

3 years ago

A student repeats the process of taking blocks out of a bag and replacing them 100 times. A green block is drawn 67 times. What

Alexeev081 [22]

Using it's concept, it is found that a good estimate for the probability of drawing out a green block from the bag is of 0.67 = 67%.

<h3>What is a probability?</h3>

A probability is given by the <u>number of desired outcomes divided by the number of total outcomes</u>.

In this problem, to get a good estimate, we get the probability taking the outcomes from the sample, that is, 67 green blocks out of 100 blocks, hence:

p = 67/100 = 0.67.

A good estimate for the probability of drawing out a green block from the bag is of 0.67 = 67%.

More can be learned about probabilities at brainly.com/question/14398287

5 0

2 years ago

A 5-card hand is dealt from a perfectly shuffled deck. Define the events: A: the hand is a four of a kind (all four cards of one

TiliK225 [7]

In a hand of 5 cards, you want 4 of them to be of the same rank, and the fifth can be any of the remaining 48 cards. So if the rank of the 4-of-a-kind is fixed, there are $\binom44\binom{48}1=48$ possible hands. To account for any choice of rank, we choose 1 of the 13 possible ranks and multiply this count by $\binom{13}1=13$ . So there are 624 possible hands containing a 4-of-a-kind. Hence A occurs with probability

$\dfrac{\binom{13}1\binom44\binom{48}1}{\binom{52}5}=\dfrac{624}{2,598,960}\approx0.00024$

There are 4 aces in the deck. If exactly 1 occurs in the hand, the remaining 4 cards can be any of the remaining 48 non-ace cards, contributing $\binom41\binom{48}4=778,320$ possible hands. Exactly 2 aces are drawn in $\binom42\binom{48}3=103,776$ hands. And so on. This gives a total of

$\displaystyle\sum_{a=1}^4\binom4a\binom{48}{5-a}=886,656$

possible hands containing at least 1 ace, and hence B occurs with probability

$\dfrac{\sum\limits_{a=1}^4\binom4a\binom{48}{5-a}}{\binom{52}5}=\dfrac{18,472}{54,145}\approx0.3412$

The product of these probability is approximately 0.000082.

A and B are independent if the probability of both events occurring simultaneously is the same as the above probability, i.e. $P(A\cap B)=P(A)P(B)$ . This happens if

the hand has 4 aces and 1 non-ace, or
the hand has a non-ace 4-of-a-kind and 1 ace

The above "sub-events" are mutually exclusive and share no overlap. There are 48 possible non-aces to choose from, so the first sub-event consists of 48 possible hands. There are 12 non-ace 4-of-a-kinds and 4 choices of ace for the fifth card, so the second sub-event has a total of 12*4 = 48 possible hands. So $A\cap B$ consists of 96 possible hands, which occurs with probability