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3 years ago

I nedd help can anyone ansewer fast please

Mathematics

1 answer:

pashok25 [27]3 years ago

3 0

18. just subtract 49.6 from 54. 54 - 49.6=4.4 He most lose 4.4 pounds.
19. all you need to do is add 4.6 to 83. 83+4.6=87.6 She will weigh 87.6 pounds
20. divide 910 by 7. 910÷7=130 The dog ate 130 grams per day
21. multiple 4.95 by 5. 4.95×5=24.75 Her total bill was 24.75
22. subtract 12.96 from 20. 20-12.96=7.04 Her change was 7.04

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Show that sec^2x+csc^2x/sec^2xcsc^2x

professor190 [17]

Answer:

$\frac{1}{csc^{2}(x)}+\frac{1}{sec^{2}(x)}$

Step-by-step explanation:

The image below shows step-by-step on how to solve it.

<em>Hope this helps! </em>:)

4 0

3 years ago

Help Please Will Give Brainliest

il63 [147K]

31 I think I’m not sure may be wrong

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3 years ago

Read 2 more answers

Most people Just most people in the nine states use feet in miles to measure distance in there every day lives why do you scient

babunello [35]

The third answer should be the right one

8 0

3 years ago

Two cards are drawn with replacement, one after the other, and the outcomes recorded. What is the probability that at least one

babymother [125]

Answer:

The probability is 0.231

Step-by-step explanation:

In this question, we are tasked with calculating the probability that out of two cards drawn with replacement from a pack of cards , at least one of the two is a face card.

In a standard pack of cards, there are 52 cards.

Now let’s calculate the number of face cards we have.

There are 12 face cards in a pack of cards. There are 4 of each of Jack, Queen or King

The probability of drawing a face card is 12/52 = 3/13

Now, we want to calculate the probability of at least one of the cards drawn are face cards.

Mathematically, what we are needing to calculate is

Probability of drawing a single face card or probability of the two being a face card

Now let’s calculate.

The probability of the first card drawn being a face card is 3/13. The probability of the second card not being a face card will be 1-3/13 = 10/13

The joint probability under this first set of condition will be 3/13 * 10/13 = 30/169

Now, for the second condition, here we are considering that both cards drawn are face cards. What we have is 3/13 * 3/13 = 9/169

Now in probabilities, whenever we are having the statement or, we add the probabilities together

Thus, the probability here would be 30/169 + 9/169 = 39/169 = 3/13

In decimal this will be 0.231

8 0

2 years ago

A box of donuts containing 6 maple bars, 3 chocolate donuts, and 3 custard filled donuts is sitting on a counter in a work offic

Svetlanka [38]

Probabilities are used to determine the chances of selecting a kind of donut from the box.

The probability that Warren eats a chocolate donut, and then a custard filled donut is 0.068

The given parameters are:

$\mathbf{Bars = 6}$

$\mathbf{Chocolate = 3}$

$\mathbf{Custard= 3}$

The total number of donuts in the box is:

$\mathbf{Total= 6 + 3 + 3}$

$\mathbf{Total= 12}$

The probability of eating a chocolate donut, and then a custard filled donut is calculated using:

$\mathbf{Pr = \frac{Chocolate}{Total}\times \frac{Custard}{Total-1}}$

So, we have:

$\mathbf{Pr = \frac{3}{12}\times \frac{3}{12-1}}$

Simplify

$\mathbf{Pr = \frac{3}{12}\times \frac{3}{11}}$

Multiply

$\mathbf{Pr = \frac{9}{132}}$

Divide

$\mathbf{Pr = 0.068}$

Hence, the probability that Warren eats a chocolate donut, and then a custard filled donut is approximately 0.068