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2 years ago

PLEASE HELP AND SHOW WORK IF YOU CAN

Mathematics

1 answer:

Murljashka [212]2 years ago

3 0

Answer:

c) XY > RS

= true

Step-by-step explanation:

Firstly know the meaning of each sign given

> means is greater than

= means is equal to

a)XZ > ST

XZ = 8

ST = 11 - greater

= false

b) XY = RS

Due to the angles presented, we conclude that the greater the angle, the longer the unmeasured line.

XY is longer than RS

= false

c) XY > RS

= true

d) RS > XY

RS is shorter than XY hence not greater.

= false

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a certain pharmaceutical company know that, on average 4% of a certain type of pill has an ingredient that is below the minimum

FinnZ [79.3K]

Using the binomial distribution, it is found that there is a 0% probability that fewer that 5 in a sample of 20 pills will be acceptable.

For each pill, there are only two possible outcomes, either it is acceptable, or it is not. The probability of a pill being acceptable is independent of any other pill, which means that the binomial distribution is used to solve this question.

Binomial probability distribution

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$C_{n,x} = \frac{n!}{x!(n-x)!}$

The parameters are:

x is the number of successes.

n is the number of trials.

p is the probability of a success on a single trial.

In this problem:

The sample has 20 pills, hence $n = 20$ .

100 - 4 = 96% are acceptable, hence $p = 0.96$

The probability that fewer that 5 in a sample of 20 pills will be acceptable is:

$P(X < 5) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4)$

In which

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 0) = C_{20,0}.(0.96)^{0}.(0.04)^{20} = 0$

$P(X = 1) = C_{20,1}.(0.96)^{1}.(0.04)^{19} = 0$

$P(X = 2) = C_{20,2}.(0.96)^{2}.(0.04)^{18} = 0$

$P(X = 3) = C_{20,3}.(0.96)^{3}.(0.04)^{17} = 0$

$P(X = 4) = C_{20,4}.(0.96)^{4}.(0.04)^{16} = 0$

0% probability that fewer that 5 in a sample of 20 pills will be acceptable.

A similar problem is given at brainly.com/question/24863377

8 0

2 years ago

Do the ratios 30/24 and 20/16 form a proportion?

BartSMP [9]

Answer:

Yes.

Step-by-step explanation:

Reduce both fractions.

30/24 = 5/4
20/16= 5/4

6 0

3 years ago

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

9. Sonya opened a savings account with and deposits each week. Brad opened a savings account with and contributes each week. Aft

Lapatulllka [165]

It will take 6 weeks for Brad to get twice the amount as Sonya.
Sonya will have $260 dollars and Brad will have $360

8 0

2 years ago

Based on the figure, which statement provides enough information to conclude that line ris

GREYUIT [131]

Answer:

m<6 = 90°

Step-by-step explanation:

If lines are perpendicular, the angles form by their intersection measure 90°.

Answer: m<6 = 90°

4 0

1 year ago

PLEASE HELP AND SHOW WORK IF YOU CAN​

PLEASE HELP AND SHOW WORK IF YOU CAN