Math help please I need it ASAP

Need help with 9 please

Debora [2.8K]

Answer:

x = 5

y = 1

Step-by-step explanation:

So move 2y from the second equation to the other side of the equals sign

2x +3y = 13

x = 3 + 2y

From here substitute X into the first equation

2(3 + 2y) + 3y = 13

Distribute

6 + 4y +3y = 13

Combine like terms and solve for y

7y = 7

y = 1

insert result into the original equation.

x-2(1) = 3

x - 2 = 3

x = 5

insert into other equation to check work

2(5) + 3(1) = 13

10 + 3 = 13 correct

6 0

3 years ago

Ken is thinking of a number. nine more than the product of 4 and the number is 73. Find Kens number.

rewona [7]

I got 16 for this one

5 0

3 years ago

Read 2 more answers

The stairs leading from the ground to the entrance of a plane forms a right triangle with the ground. If the distance of the sta

9966 [12]

Answer:B

Step-by-step explanation:

3 0

2 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

(c) Here, also 8 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 = 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

2 years ago

Why isn't a dilation considered a rigid transformation?

dexar [7]

It's not rigid because dilations (scale factor not equal to 1) change the length of the segments, or the distances between the points. You'll get a similar figure but it won't be congruent. For example, if the scale factor is 3, then the distances will be three times as large; or the lengths will be 3 times as long.

To be "rigid", the lengths must be kept the same. In contrast, a reflection is rigid because the distances are kept the same. The only thing changing is the orientation (clockwise to counter-clockwise, or vice versa).

7 0

3 years ago