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3 years ago

The mass of Earth is made up of these parts.

Mathematics

2 answers:

Kamila [148]3 years ago

6 0

Answer:

Edg.enuity answers

Step-by-step explanation:

1. Atmosphere- 8.90118

2. Oceans- 2.44346

3. Crust- 4.53786

4. Inner Core- 1.68861

5. Outer Core- 3.20268

6. Mantle- 7.05637

Kipish [7]3 years ago

4 0

Answer: 1. Atmosphere, 2. Oceans, 3. Crust, 4. Inner core, 5. Outer core, 6. Mantle

Step-by-step explanation: correct on e d g e n u i t y

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Example:The missing value is x than Sin 7 / 18

Darya [45]

Answer:

44.9

Step-by-step explanation:

12/17 = PQ/PR

(sin^-1)(12/17) = 44.9

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3 years ago

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A water tank is being filled with water. Before water is pumped in, the water level in the tank is 11 inches high. After 20 minu

serg [7]

Answer:

(58-11) divided by 20

Step-by-step explanation:

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3 years ago

Given the equation y= 1/2x + 9, what is the y-intercept?

In-s [12.5K]

Answer:

Step-by-step explanation:

Y intercept when x = 0

Y = 9

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2 years ago

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19125 people, exactly 75%, what is the total population of the town

Pepsi [2]

Answer:

25500 people

Step-by-step explanation:

We are given;

Number of people is 19125

The percentage representing the number is 75%

We need to determine the total population;

The total population will be 100%

Therefore;

If 75% = 19125 people

Then, 100% will represent;

= (19125 ÷ 75) × 100

= 255 × 100

= 25500 people

Therefore, the total population of the town is 25500 people

8 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

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