Which of the following is NOT an example of matter?

Please help me....))

miskamm [114]

The first picture is for number 1 and the second one for number 2.

8 0

3 years ago

If cos(x)=1/4 what is sin(x) and tan(x)

anygoal [31]

You can use the identity
cos(x)² +sin(x)² = 1
to find sin(x) from cos(x) or vice versa.

(1/4)² +sin(x)² = 1
sin(x)² = 1 - 1/16
sin(x) = ±(√15)/4

Then the tangent can be computed as the ratio of sine to cosine.
tan(x) = sin(x)/cos(x) = (±(√15)/4)/(1/4)
tan(x) = ±√15

There are two possible answers.
In the first quadrant:
sin(x) = (√15)/4
tan(x) = √15

In the fourth quadrant:
sin(x) = -(√15)/4
tan(x) = -√15

3 0

3 years ago

Can i please get your help , it will mean a lot if you actually do and luvya so for helping genuinely :)

Art [367]

Answer:29

Step-by-step explanation:

Opposite angles in a quadrilateral (the trapezium inside the circle) must add up to 180, so 180-92 = 88.

So DAB = 88

Angles on a straight line add up to 180, so to find t, you have to 180 - 88 - 63 = 29

So t is 29.

Hope that helps!

8 0

3 years ago

Which of the follow expressions are equivalent to (3^4)^-2 • (3^5)^3

Fiesta28 [93]

Answer:

3^7

Step-by-step explanation:

3 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

(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

3 years ago