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3 years ago

Using the digits 1 to 9 at most one time each, fill in the boxes to make the equality true:

Mathematics

1 answer:

spin [16.1K]3 years ago

8 0

Answer:

(4+5-2+1)- (7+3-8+6) = 0

Step-by-step explanation:

you look for what numbers when subtracted equal zero (in this case each of the parentheses is 8) in each of the parentheses and combine them at the end to form the final equation.

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What are the steps to this?

Evgen [1.6K]

Answer:

B=112

Step-by-step explanation:

Angle B is on the same line as angle A(68). They would need to combine for a total of 180 degrees. So 68-180=112

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3 years ago

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Write the fraction as the product of a whole number and. Unit fraction 2/5

zavuch27 [327]

Answer:

I don't know exactly what you mean but I think the answer is .4

Step-by-step explanation:

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3 years ago

The equilibrium constant for this system is

blsea [12.9K]

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a. 9.2 x 10-2 M

Step-by-step explanation:

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3 years ago

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Harman [31]

Answer:

Harper is 200 yards away from her home if the map shows 3 inches

Step-by-step explanation:

50 yards = 3/4 inch

3 inches = 3/1

3/1 x 4/4 = 12/4

12/4 = 3 inches

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I hope this helped and if it did I would appreciate it if you marked me Brainliest. Thank you and have a nice day!

6 0

2 years ago

A computer can be classified as either cutting dash edge or ancient. Suppose that 94% of computers are classified as ancient.

taurus [48]

Answer:

(a) 0.8836

(b) 0.6096

(c) 0.3904

Step-by-step explanation:

We are given that a computer can be classified as either cutting dash edge or ancient. Suppose that 94% of computers are classified as ancient.

(a) Two computers are chosen at random.

The above situation can be represented through Binomial distribution;

$P(X=r) = \binom{n}{r}p^{r} (1-p)^{n-r} ; x = 0,1,2,3,.....$

where, n = number of trials (samples) taken = 2 computers

r = number of success = both 2

p = probability of success which in our question is % of computers

that are classified as ancient, i.e; 0.94

LET X = Number of computers that are classified as ancient

So, it means X ~ $Binom(n=2, p=0.94)$

Now, Probability that both computers are ancient is given by = P(X = 2)

P(X = 2) = $\binom{2}{2}\times 0.94^{2} \times (1-0.94)^{2-2}$

= $1 \times 0.94^{2} \times 1$

= 0.8836

(b) Eight computers are chosen at random.

The above situation can be represented through Binomial distribution;

$P(X=r) = \binom{n}{r}p^{r} (1-p)^{n-r} ; x = 0,1,2,3,.....$

where, n = number of trials (samples) taken = 8 computers

r = number of success = all 8

p = probability of success which in our question is % of computers

that are classified as ancient, i.e; 0.94

LET X = Number of computers that are classified as ancient

So, it means X ~ $Binom(n=8, p=0.94)$

Now, Probability that all eight computers are ancient is given by = P(X = 8)

P(X = 8) = $\binom{8}{8}\times 0.94^{8} \times (1-0.94)^{8-8}$

= $1 \times 0.94^{8} \times 1$

= 0.6096

The above situation can be represented through Binomial distribution;

$P(X=r) = \binom{n}{r}p^{r} (1-p)^{n-r} ; x = 0,1,2,3,.....$

where, n = number of trials (samples) taken = 8 computers

r = number of success = at least one

p = probability of success which is now the % of computers

that are classified as cutting dash edge, i.e; p = (1 - 0.94) = 0.06

LET X = Number of computers classified as cutting dash edge

So, it means X ~ $Binom(n=8, p=0.06)$

Now, Probability that at least one of eight randomly selected computers is cutting dash edge is given by = P(X $\geq$ 1)

P(X $\geq$ 1) = 1 - P(X = 0)

= $1 - \binom{8}{0}\times 0.06^{0} \times (1-0.06)^{8-0}$

= $1 - [1 \times 1 \times 0.94^{8}]$

= 1 - $0.94^{8}$ = 0.3904

Here, the probability that at least one of eight randomly selected computers is cutting dash edge is 0.3904 or 39.04%.

For any event to be unusual it's probability is very less such that of less than 5%. Since here the probability is 39.04% which is way higher than 5%.

So, it is not unusual that at least one of eight randomly selected computers is cutting dash edge.

7 0

3 years ago