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2 years ago

Find an ordered pair that satisfies the equation 81x + 18y = 9

Mathematics

2 answers:

ch4aika [34]2 years ago

4 0

We can solve using y = 0 and solving for x.

81x + 18(0) = 9

81x = 9

x = 1/9

Therefore, the point (1/9, 0) sastisfies the equation.

Best of Luck!

Murrr4er [49]2 years ago

4 0

Find either the x intercept or y intercept:

To find the y intercept, Set x to 0 and solve for y:

81(0) + 18y = 9

0 + 18y = 9

Divide both sides by 18:

Y = 9/18

Y = 1/2

One solution is (0,1/2)

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Julia has 4 gallons 2 quarts of water.Ally needs the same amount of water but only has 12 quarts.How much more water does Ally n

ozzi

Ok so what do we know :

4 quarts = 1 gal
Julia has 4gal and 2 quarts or 18 quarts
ally has 12 quarts

so if julia has 18 quarts and ally has 12 that means that ally needs 6 more quarts (or 1gal 2 quarts) so that she and julia can have the same amount

8 0

3 years ago

Read 2 more answers

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.

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2 years ago

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tensa zangetsu [6.8K]

Answer:

none

Step-by-step explanation:

he bought a bunch of chicken and watermelon so he didnt save it

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10. Solve the following pair of simultaneous equations 2 + x = y xy = 5. 1 Add file

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x = 5 and y = 3. That is the answer

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2 years ago

Please help! thanks

anygoal [31]

1 =1,1,4. 5=13,16

2 =1,4. 6=23,3

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4 =3,16.

ps , meanes fraction

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