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3 years ago

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Create three fractions whose product is -5/24. You may only use fractions between -8/9 to 8/9 and you can only use a fraction on

ce.
Find at least 2 possible combinations.

1 answer:

damaskus [11]3 years ago

8 0

Answer:

We find 3/4-5/24 = 0.54166666666 and try 2/7 x 6/8 = 0.21428571428

this is 0.069 too much we look at this as a fraction = 0.07 which is 1/10 approx of the 6/8. we then reapply 7/9 x 2/7 x 3/3 = 0.2

So we went up by 1 and can show another

2/7 x 7/10 = 2.0

So to 1dp there is lots of fractions including 2/7 x 7/9 = 2.0

But for three fractions that multiply together we can chose 5/7 x 2/6 x 6/6 = 2.0 As 5 x 2 x 6 = 60 and 60/360-102 = 60/258 = 0.20 and divided 198 =numbers 8 and 1/3 work with = 1.98/8.3 = 2.38554

We try again 1.68/8.3 = -0.2024 = 2/8 x 8/10 = 0.2

The balance is 1.3/2 x 0.8/2 = 0.65 x 0.333 = 13/20 x 1/3

3/5 x 1/3 = 0.2

You might be interested in

Consider a value to be significantly low if its z score less than or equal to minus−2 or consider a value to be significantly hi

den301095 [7]

Answer:

Test scores of 10.2 or lower are significantly low.

Test scores of 31 or higher are significantly high

Step-by-step explanation:

Z-score:

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

$\mu = 20.6, \sigma = 5.2$

Significantly low:

Z-scores of -2 or lower

So scores of X when Z = -2 or lower

$Z = \frac{X - \mu}{\sigma}$

$-2 = \frac{X - 20.6}{5.2}$

$X - 20.6 = -2*5.2$

$X = 10.2$

Test scores of 10.2 or lower are significantly low.

Significantly high:

Z-scores of 2 or higher

So scores of X when Z = 2 or higher

$Z = \frac{X - \mu}{\sigma}$

$2 = \frac{X - 20.6}{5.2}$

$X - 20.6 = 2*5.2$

$X = 31$

Test scores of 31 or higher are significantly high

6 0

3 years ago

A bag contains two six-sided dice: one red, one green. The red die has faces numbered 1, 2, 3, 4, 5, and 6. The green die has fa

gayaneshka [121]

Answer:

the probability the die chosen was green is 0.9

Step-by-step explanation:

Given that:

A bag contains two six-sided dice: one red, one green.

The red die has faces numbered 1, 2, 3, 4, 5, and 6.

The green die has faces numbered 1, 2, 3, 4, 4, and 4.

From above, the probability of obtaining 4 in a single throw of a fair die is:

P (4 | red dice) = $\dfrac{1}{6}$

P (4 | green dice) = $\dfrac{3}{6}$ = $\dfrac{1}{2}$

A die is selected at random and rolled four times.

As the die is selected randomly; the probability of the first die must be equal to the probability of the second die = $\dfrac{1}{2}$

The probability of two 1's and two 4's in the first dice can be calculated as:

= $\begin {pmatrix} \left \begin{array}{c}4\\2\\ \end{array} \right \end {pmatrix} \times \begin {pmatrix} \dfrac{1}{6} \end {pmatrix} ^4$

= $\dfrac{4!}{2!(4-2)!} ( \dfrac{1}{6})^4$

= $\dfrac{4!}{2!(2)!} \times ( \dfrac{1}{6})^4$

= $6 \times ( \dfrac{1}{6})^4$

= $(\dfrac{1}{6})^3$

= $\dfrac{1}{216}$

The probability of two 1's and two 4's in the second dice can be calculated as:

= $\begin {pmatrix} \left \begin{array}{c}4\\2\\ \end{array} \right \end {pmatrix} \times \begin {pmatrix} \dfrac{1}{6} \end {pmatrix} ^2 \times \begin {pmatrix} \dfrac{3}{6} \end {pmatrix} ^2$

= $\dfrac{4!}{2!(2)!} \times ( \dfrac{1}{6})^2 \times ( \dfrac{3}{6})^2$

= $6 \times ( \dfrac{1}{6})^2 \times ( \dfrac{3}{6})^2$

= $( \dfrac{1}{6}) \times ( \dfrac{3}{6})^2$

= $\dfrac{9}{216}$

∴

The probability of two 1's and two 4's in both dies = P( two 1s and two 4s | first dice ) P( first dice ) + P( two 1s and two 4s | second dice ) P( second dice )

The probability of two 1's and two 4's in both die = $\dfrac{1}{216} \times \dfrac{1}{2} + \dfrac{9}{216} \times \dfrac{1}{2}$

The probability of two 1's and two 4's in both die = $\dfrac{1}{432} + \dfrac{1}{48}$

The probability of two 1's and two 4's in both die = $\dfrac{5}{216}$

By applying Bayes Theorem; the probability that the die was green can be calculated as:

P(second die (green) | two 1's and two 4's ) = The probability of two 1's and two 4's | second dice)P (second die) ÷ P(two 1's and two 4's in both die)

P(second die (green) | two 1's and two 4's ) = $\dfrac{\dfrac{1}{2} \times \dfrac{9}{216}}{\dfrac{5}{216}}$

P(second die (green) | two 1's and two 4's ) = $\dfrac{0.5 \times 0.04166666667}{0.02314814815}$

P(second die (green) | two 1's and two 4's ) = 0.9

Thus; the probability the die chosen was green is 0.9

8 0

3 years ago

Use the figure to find n please.

ira [324]

Answer:

n = 5

Step-by-step explanation:

Since this is a right triangle, we can use trig

tan theta = opp /adj

tan 30 = n / 5 sqrt(3)

5 sqrt(3) tan 30 = n

5 sqrt(3) * 1/ sqrt(3) = n

5 = n

4 0

2 years ago

Read 2 more answers

19+4n<br><br> But "n=4"<br><br> IF YOU DON'T EXPLAIN OR PUT A FAKE ANSWER YOU WILL BE REPORTED

salantis [7]

Sub the number in
19 + 4(4)
= 19 + 16
= 35
multiplication is first because of bedmas

8 0

3 years ago

Read 2 more answers

Which number is IRRATIONAL ?? (if answered really fast, i will give brainiest plus 100 points for the CORRECT answer.

Delvig [45]

Answer:

C

btw Hannah is this one fine? sorry i couldn't remember which question was the chat and sorry i couldn't respond right away im not the person who is on late in the day

Step-by-step explanation:

3 0

2 years ago