All categories

2 years ago

ANYONE GOT SNAP INSTA OR DISCORD????

Mathematics

2 answers:

Dmitriy789 [7]2 years ago

7 0

Answer:

I GOT DISCORD, NO SNAP OR INSTA THO

Step-by-step explanation:

WHY? I DONT KNOW WHY IM TYPING IN CAPS BY THE WAY

Aleksandr [31]2 years ago

4 0

YESSSSSSS I HAVEEEEE DISCORDDDDD

You might be interested in

Which expression does NOT show hot Evelyn can find the volume of the box?

Pie

First one. volume is L*W*H and it should be multiplication not addition (the 6+6 on an on) one works because it is adding all the numbers

5 0

3 years ago

Read 2 more answers

What is 8.05 × 10-5 in standard notation? 0.0000805 0.00000805 805,000 8,050,000

Tanya [424]

Answer:

0.0000805

Step-by-step explanation:

$8.05*10^{-5}=\frac{8.05}{10^{5}}\\\\=\frac{8.05}{100,000}\\$

= 0.0000805

4 0

3 years ago

Read 2 more answers

What are the two solutions of 2|2y-1| + 5 = 19

Bess [88]

Answer:

y=4

y=−3

Step-by-step explanation:

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.

7 0

3 years ago

Compare 7/12 and 1/2

Klio2033 [76]

7/12 and 1/2
Get the common denominator
Multiply 1/2 by 6
7/12 and 6/12

3 0

3 years ago

Read 2 more answers

ANYONE GOT SNAP INSTA OR DISCORD???? ​

ANYONE GOT SNAP INSTA OR DISCORD????