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Sergeu [11.5K]
2 years ago
8

Which of the following lists is ordered from least to greatest?

Mathematics
1 answer:
lisabon 2012 [21]2 years ago
4 0

Answer:

-5,0,0.8,1,1 is the answer

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Please help a girl outtttt
jeyben [28]

Answer:

bottom of graph will move from (0,0) to point (1,3) after transformation

Step-by-step explanation:

given

original : f(x) = x^{2}

transformed; g(x) = (x-1)^{2} + 3

look at this way g(x) =  (x-h)^{2} + k

if (x-h), h>0, move h units to the right

if k>0, move k units up

the bottom of the graph will be at point (1,3)

3 0
2 years ago
Each marker weighs 10 g There are 8 markers in a pack. How much do all the markers weigh?
Ilya [14]
80 g 10 times 8 is 80 so the answer should be 80g
8 0
2 years ago
A card is drawn from a standard deck of 5252 playing cards. What is the probability that the card will be a heart or a face card
mr_godi [17]

Answer:

The probability that the card will be a heart or a face card is P=0.4231.

Step-by-step explanation:

We have a standard deck of 52 cards.

In this deck, we have 13 cards that are a heart.

We also have a total of 12 face cards (4 per each suit). So there are 3 face cards that are also a heart.

To calculate the probability that a card be a heart or a face card, we sum the probability of a card being a heart and the probability of it being a face card, and substract the probability of being a heart AND a face card.

We can express that as:

P(H\,or\,F)=P(H)+P(F)-P(H\&F)\\\\P(H\,or\,F)=13/52+12/52-3/52=22/52=0.4231

4 0
3 years ago
Suppose we roll a fair six-sided die and sum the values obtained on each roll, stopping once our sum exceeds 307. Approximate th
SOVA2 [1]

Answer:

There is a probability of P=0.94 that at least 81 rolls are needed to get the sum of 307.

Step-by-step explanation:

The roll of a six-sided die sum has this parameters:

Mean: 3.5

Standard deviation: 1.7

If the die is rolled 81 times, the distribution of the sum of the 81 rolls will have the following parameters:

\mu=n*M=81*3.5=283.5\\\\\sigma=\sqrt{n}s=\sqrt{81}*1.7=9*1.7=15.3

Note: the variance of the sum of random variables is equal to the sum of the variance of the individual variables.

Then, we can calculate the probabilties that the sum of 81 rolls is lower than 307.

z=\frac{x-\mu}{\sigma}=\frac{307-283.5}{15.3}=  \frac{23.5}{15.3}= 1.536\\\\\\P(x

There is 94% of chances that the sum of 81 rolls is lower than 307, so there is a probability of P=0.94 that at least 81 rolls are needed to get the sum of 307.

5 0
3 years ago
Simply the expressions:
Marta_Voda [28]
= 11x
= 13x
= 45x
= 16x + 8y
= 21x + 35y
= 11x
8 0
3 years ago
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