Brian compared the decimals 66.10, 21.10, and 66.11.

An eperiment conducted found that is the last 25 days a blue pen 5 times. There are 6 blue pens and 14 black pens. What is the e

Bad White [126]

6 out of 14 i think try it

3 0

2 years ago

Please help!!! I would like an explanation along with your answer. Thanks

Tpy6a [65]

Answer:

11.2 feet

Step-by-step explanation:

The two furthest corners of the bed are at (8, 6) and (5, 10). To determine which is the furthest from the origin, use the distance formula.

d = √((x₂ − x₁)² + (y₂ − y₁)²)

d = √((8 − 0)² + (6 − 0)²)

d = 10

d = √((x₂ − x₁)² + (y₂ − y₁)²)

d = √((5 − 0)² + (10 − 0)²)

d = 5√5

d ≈ 11.2

Therefore, the corner at (5, 10) is the furthest from the origin, about 11.2 feet away.

8 0

3 years ago

If the triangles are similar, what is the value of x?

Sav [38]

Answer:

the answer to your question is D. d=54

4 0

3 years ago

Combine the like terms to create an equivalent expression. Pleaseee helppp ASAP thankss

Nonamiya [84]

The answer is 2z+3. The like terms are 5z and -3z so you can subtract them and get 2z. Thus, the answer is 2z+3

3 0

3 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

(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