Please help with number 6 ( at the top)

In one year the perseid metor shower had a metor appear every 1/5 5 minutes on average That same year the Leonid metor shower ha

olga55 [171]

In one year, the Perseid meteor shower had a meteor appear every 1 1/5 minutes on average. That same year, the Leonid meteor shower had a meteor appear every 4 2/3 minutes on average. How many more meteors fell during the Perseid meteor shower?

Answer:

325,452 meteors

Step-by-step explanation:

We all know that :

We have 365 days in a year and in each day there are 24 hours, Likewise an hour has 60 minutes

SO, the total number of minutes in a year is :

365 × 24 × 60 = 525600 minutes

For Perseid meteor:

$1 + \frac{1}{5} \\ \\ = 1 + 0.2 \\ \\ = 1.2$

So one meteor fall at 1.2 minutes interval Thus, in a year, we will have :

1 meteor - 1.2 minutes.

x meteor - 525600 minutes.

1.2x = 525600

$x = \frac{525600}{1.2}$

x = 438000

∴ 438,000 meteors fell during the Perseid shower.

For Leonid meteor shower:

$4 + \frac{2}{3} = 4.67$

So one meteor fall at 4.67 minutes interval Thus, in a year, we will have :

1 meteor - 4.67 minutes.

x meteors - 525600 minutes.

$4.67x = 525600$

$x = \frac{525600}{4.67}$

x = 112548

112,548 meteors fell during the Leonid Shower.

Finally , the numbers of more meteors that fell during the Perseid meteor shower is calculated by the difference in the number of Perseid meteor shower and Lenoid meteor shower. i.e

438,000 - 112,548 = 325,452

325,452 more meteors fell during the Perseid meteor shower

5 0

3 years ago

Dont tell me the Answer but walk me through the steps plz

sveticcg [70]

Answer:

Find the multiplies of 2

7 0

3 years ago

Please give a good answer :)

Alex

Answer:

P(4≤x≤7) = 2/3

Step-by-step explanation:

We'll begin by obtaining the sample space (S) i.e possible outcome of rolling both dice at the same time. This is illustrated below:

1,1 1,2 1,3 1,4 1,5 1,6

2,1 2,2 2,3 2,4 2,5 2,6

3,1 3,2 3,3 3,4 3,5 3,6

4,1 4,2 4,3 4,4 4,5 4,6

5,1 5,2 5,3 5,4 5,5 5,6

6,1 6,2 6,3 6,4 6,5 6,6

Adding the outcome together, the sample space (S) becomes:

2 3 4 5 6 7

3 4 5 6 7 8

4 5 6 7 8 9

5 6 7 8 9 10

6 7 8 9 10 11

7 8 9 10 11 12

Next, we shall obtain the event of 4≤x≤7. This is illustrated below:

4 5 6 7

Finally, we shall determine P(4≤x≤7). This can be obtained as follow:

Element in the sample space, n(S) = 36

Element in 4≤x≤7, n(4≤x≤7) = 24

Probability of 4≤x≤7, P(4≤x≤7) = ?

P(4≤x≤7) = n(4≤x≤7) / nS

P(4≤x≤7) = 24/36

P(4≤x≤7) = 2/3

4 0

3 years ago

I NEED HELP PLZ AHHHH

rjkz [21]

With what? I might be able to help

3 0

3 years ago

Classify ABC by its sides. Then determine whether it is a right triangle.

m_a_m_a [10]

Answer:

∴Given Δ ABC is not a right-angle triangle

a= AB = √45 = 3√5

b = BC = 12

c = AC = √45 = 3√5

Step-by-step explanation:

Given vertices are A(3,3) and B(6,9)

$AB = \sqrt{x_{2}-x_{1} )^{2} +(y_{2}-y_{1} )^{2} }$

AB = $\sqrt{(9-3)^{2}+(6-3)^{2} } = \sqrt{6^{2}+3^{2} } =\sqrt{45}$

Given vertices are B(6,9) and C( 6,-3)

$B C = \sqrt{x_{2}-x_{1} )^{2} +(y_{2}-y_{1} )^{2} }$

= $\sqrt{(-3-9)^{2}+(6-6)^{2} } =\sqrt{12^{2} } = 12$

BC = 12

Given vertices are A(3,3) and C( 6,-3)

$AC = \sqrt{(6-3)^{2}+(-3-3)^{2} } = \sqrt{9+36} = \sqrt{45}$

AC² = AB²+BC²

45 = 45+144

45 ≠ 189

∴Given Δ ABC is not a right angle triangle

5 0

3 years ago