Use the Venn diagram to calculate probabilities. Which probability is correct?

Stuck on this hope someone can help me

Murljashka [212]

Answer:

all are correct except the third one. the formula for that one should be v = a³ since it's a cube

Step-by-step explanation:

3 0

3 years ago

12 A pilot needs to know the maximum height an aircraft can fly at. The cabin has been tested and is safe to a height of 15679 m

abruzzese [7]

15.6 kilometers rounded down because it is not safe to go above what has been approved

7 0

2 years ago

Help please :) about to post more too

yKpoI14uk [10]

Answer:

Lets say test tubes = t, and beakers = b

1 pack of (t) is $4 less than 1 pack of (b)

Since i have no prior information we are going to use variables for this equation:

1t (1 pack of test tubes) is $4 less than 1b (1 set of beakers)

so to quantify the equation, we have 8t and 12b.

if b is a number that IS quantifiable such as $5 we can easily figure out this answer.

Lets use and example that 1 set of beakers is $8, if we multiply $8 by 12 (the number of sets of beakers), we get: 96

Using the same example, if 1t is $4 less than 1b than 1t = $4. So, if we multiply $4 by 8 (the amount of packs of test tubes), we get: 32

If you take both of those numbers: 96, and 32 and you divide them you get 3. so that means that 1t = 3b

Answer = 1t = 3b

This may not be correct due to the little information that i got however i hope that, that works out for you :)

5 0

3 years ago

Each year for 4 years, a farmer increased the number of trees in a certain orchard by of the number of trees in the orchard the

Neko [114]

Answer:

The number of trees at the begging of the 4-year period was 2560.

Step-by-step explanation:

Let’s say that x is number of trees at the begging of the first year, we know that for four years the number of trees were incised by 1/4 of the number of trees of the preceding year, so at the end of the first year the number of trees was $x+\frac{1}{4} x=\frac{5}{4} x$ , and for the next three years we have that

Start End

Second year $\frac{5}{4}x$ -------------- $\frac{5}{4}x+\frac{1}{4}(\frac{5}{4}x) =\frac{5}{4}x+ \frac{5}{16}x=\frac{25}{16}x=(\frac{5}{4} )^{2}x$

Third year $(\frac{5}{4} )^{2}x$ ------------- $(\frac{5}{4})^{2}x+\frac{1}{4}((\frac{5}{4})^{2}x) =(\frac{5}{4})^{2}x+\frac{5^{2} }{4^{3} } x=(\frac{5}{4})^{3}x$

Fourth year $(\frac{5}{4})^{3}x$ -------------- $(\frac{5}{4})^{3}x+\frac{1}{4}((\frac{5}{4})^{3}x) =(\frac{5}{4})^{3}x+\frac{5^{3} }{4^{4} } x=(\frac{5}{4})^{4}x.$

So the formula to calculate the number of trees in the fourth year is

$(\frac{5}{4} )^{4} x,$ we know that all of the trees thrived and there were 6250 at the end of 4 year period, then

$6250=(\frac{5}{4} )^{4}x$ ⇒ $x=\frac{6250*4^{4} }{5^{4} }= \frac{10*5^{4}*4^{4} }{5^{4} }=2560.$

Therefore the number of trees at the begging of the 4-year period was 2560.

7 0

3 years ago

An unfair die is twice as likely to roll an even number than an odd number. what is the probability of it rolling an odd number?

Anarel [89]

The probability is always expressed as a part of the whole. So, it can be in terms of fraction or percentage. The numerator is the number of possible events while the denominator is the number of total events. Let x be the number of times you roll an odd number. So, the number of times you roll an even number would be 2x. The total rolls would then be x + 2x. Thus, the probability of rolling an odd number is

Probability = x/(2x + x) = x/3x = 1/3

3 0

3 years ago