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3 years ago

Hello i need some help with these I'm not the best at math thx

Mathematics

2 answers:

lana [24]3 years ago

8 0

The first answer is D and the second answer is B.

saul85 [17]3 years ago

7 0

Answer For the First One:

D, 51.3%

Step-by-step explanation:

The total number of students that ordered popcorn was 80 and 80 is 51.3% of 156 (the total number of students)!

Answer For the Second One:

C, It was reflected over the X-axis

Explanation:

If you just flip the DEF backwards over the x-axis it makes the GHI shape!

Hope this helps! :)

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According to one cosmological theory, there were equal amounts of the two uranium isotopes 235U and 238U at the creation of the

FromTheMoon [43]

Answer:

6 billion years.

Step-by-step explanation:

According to the decay law, the amount of the radioactive substance that decays is proportional to each instant to the amount of substance present. Let $P(t)$ be the amount of $^{235}U$ and $Q(t)$ be the amount of $^{238}U$ after $t$ years.

Then, we obtain two differential equations

$\frac{dP}{dt} = -k_1P \quad \frac{dQ}{dt} = -k_2Q$

where $k_1$ and $k_2$ are proportionality constants and the minus signs denotes decay.

Rearranging terms in the equations gives

$\frac{dP}{P} = -k_1dt \quad \frac{dQ}{Q} = -k_2dt$

Now, the variables are separated, $P$ and $Q$ appear only on the left, and $t$ appears only on the right, so that we can integrate both sides.

$\int \frac{dP}{P} = -k_1 \int dt \quad \int \frac{dQ}{Q} = -k_2\int dt$

which yields

$\ln |P| = -k_1t + c_1 \quad \ln |Q| = -k_2t + c_2$ ,

where $c_1$ and $c_2$ are constants of integration.

By taking exponents, we obtain

$e^{\ln |P|} = e^{-k_1t + c_1} \quad e^{\ln |Q|} = e^{-k_12t + c_2}$

Hence,

$P = C_1e^{-k_1t} \quad Q = C_2e^{-k_2t}$ ,

where $C_1 := \pm e^{c_1}$ and $C_2 := \pm e^{c_2}$ .

Since the amounts of the uranium isotopes were the same initially, we obtain the initial condition

$P(0) = Q(0) = C$

Substituting 0 for $P$ in the general solution gives

$C = P(0) = C_1 e^0 \implies C= C_1$

Similarly, we obtain $C = C_2$ and

$P = Ce^{-k_1t} \quad Q = Ce^{-k_2t}$

The relation between the decay constant $k$ and the half-life is given by

$\tau = \frac{\ln 2}{k}$

We can use this fact to determine the numeric values of the decay constants $k_1$ and $k_2$ . Thus,

$4.51 \times 10^9 = \frac{\ln 2}{k_1} \implies k_1 = \frac{\ln 2}{4.51 \times 10^9}$

and

$7.10 \times 10^8 = \frac{\ln 2}{k_2} \implies k_2 = \frac{\ln 2}{7.10 \times 10^8}$

Therefore,

$P = Ce^{-\frac{\ln 2}{4.51 \times 10^9}t} \quad Q = Ce^{-k_2 = \frac{\ln 2}{7.10 \times 10^8}t}$

We have that

$\frac{P(t)}{Q(t)} = 137.7$

Hence,

$\frac{Ce^{-\frac{\ln 2}{4.51 \times 10^9}t} }{Ce^{-k_2 = \frac{\ln 2}{7.10 \times 10^8}t}} = 137.7$

Solving for $t$ yields $t \approx 6 \times 10^9$ , which means that the age of the universe is about 6 billion years.

5 0

3 years ago

Nine times the difference of a number and one is equal to seven less than two times the number

Oksana_A [137]

9*(x-1)=2x-7
First get rid of your parenthesis by distributing.
9x-9=2x-7
Then all you have to do is isolate the variable
9x=2x+2
7x=2
so x= 2/7

8 0

2 years ago

a parking garage has 6 levels. Each level has 15 row. Each row has the same number of parking spaces. there are 2,250 parking sp

Mandarinka [93]

To start with, we know there’s a definite amount of parking spaces in all(2,250) and we also know each level(6 in all) had 15 rows of parking spaces.

Equation

A = Parking spots per row

A = 2,250 Divided by(6 levels times 15 rows)

A= 2,250 Divided by 90

A = 25

The correct answer is 25 parking spots per row

To check this answer we can do

25 spots per row, Times 15 rows per level, Times the total amount of levels (6)

= the total amount of parking spots

25*15*6= 2,250

True

This confirms this answer as correct. Hope this helps.

6 0

2 years ago

Six different colored dice are rolled. Of interest is the number of dice that show a one. In words, define the random variable X

olga nikolaevna [1]

Answer:

Step-by-step explanation:

Six dice (each of a different color) are rolled. Since the number of times the 6 dice were rolled isn't stated, take it to be 1.

Hence, 6 dice are rolled at once for this experiment.

Of interest is the number of dice that show a "one". In other words, the variable in question (X) is:

The number of dots that show on the upward face of the rolled dice.

The values that X may take on are:

1 2 3 4 5 6

On average, how many dice would you expect to show a one?

One die.

How is this gotten? By finding the probability that a one or one dot appears when the 6 dice are rolled at once. Since there are 6 dice in number, and each die has the same 6 faces containing dots, the probability of getting a one is 1/6. In this case, one out of 6 dice is expected to show one dot.

Find the probability that all six dice show a dot in just one toss.

Logically, this probability is going to be very small! It is almost impossible for all 6 dice to land on the same face in just a single toss. In other words, expect many decimal places in the probability figure.

1/6 divided by 6 = 0.167/6 = 0.028 approximated to three decimal places.

This would also represent the probability that 2 dots, 3 dots, or any other number of dots appears on all dice in one toss!

Is it more likely (in this experiment) that 3 or 4 dice will show a one (a single dot)?

The answer is yes! It is more likely that 3 or 4 dice (instead of all 6) will show a one or will show the same number of dots.

Kudos!

7 0

3 years ago

A company hires five people for the same job for one week. The amount that each person is paid

dangina [55]

Answer:

Step-by-step explanation:

4 0

2 years ago

Read 2 more answers