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Maksim231197 [3]
3 years ago
5

Need answers pls I am struggling!

Mathematics
1 answer:
Lady bird [3.3K]3 years ago
6 0

Answer:

(1)....27

(2)...30

ans...............

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1
Dafna1 [17]
Your problem is too spaced out too understand! Next time make sure he spacing is more understandable and readable! sorry
5 0
3 years ago
√₂
snow_lady [41]

Answer:

$2000 was invested at 5% and $5000 was invested at 8%.

Step-by-step explanation:

Assuming the interest is simple interest.

<u>Simple Interest Formula</u>

I = Prt

where:

  • I = interest earned.
  • P = principal invested.
  • r = interest rate (in decimal form).
  • t = time (in years).

Given:

  • Total P = $7000
  • P₁ = principal invested at 5%
  • P₂ = principal invested at 8%
  • Total interest = $500
  • r₁ = 5% = 0.05
  • r₂ = 8% = 0.08
  • t = 1 year

Create two equations from the given information:

\textsf{Equation 1}: \quad \sf P_1+P_2=7000

\textsf{Equation 2}: \quad \sf P_1r_1t+P_2r_2t=I\implies 0.05P_1+0.08P_2=500

Rewrite Equation 1 to make P₁ the subject:

\implies \sf P_1=7000-P_2

Substitute this into Equation 2 and solve for P₂:

\implies \sf 0.05(7000-P_2)+0.08P_2=500

\implies \sf 350-0.05P_2+0.08P_2=500

\implies \sf 0.03P_2=150

\implies \sf P_2=\dfrac{150}{0.03}

\implies \sf P_2=5000

Substitute the found value of P₂ into Equation 1 and solve for P₁:

\implies \sf P_1+5000=7000

\implies \sf P_1=7000-5000

\implies \sf P_1 = 2000

$2000 was invested at 5% and $5000 was invested at 8%.

Learn more about simple interest here:

brainly.com/question/27743947

brainly.com/question/28350785

5 0
1 year ago
425x4 solve using partial product method
cluponka [151]
The answer to the question is 1700
7 0
3 years ago
USA Today reports that about 25% of all prison parolees become repeat offenders. Alice is a social worker whose job is to counse
pychu [463]

Answer:

a) P(X=0)=(4C0)(0.75)^0 (1-0.75)^{4-0}=0.0039  

P(X=1)=(4C1)(0.75)^1 (1-0.75)^{4-1}=0.0469  

P(X=2)=(4C2)(0.75)^2 (1-0.75)^{4-2}=0.211  

P(X=3)=(4C3)(0.75)^2 (1-0.75)^{4-3}=0.422  

P(X=4)=(4C4)(0.75)^2 (1-0.75)^{4-4}=0.316

b) E(X) = np = 4*0.75=3

c) Sd(X) =\sqrt{np(1-p)}=\sqrt{4*0.75*(1-0.75)}=0.866

d) P(X \geq 3) \geq 0.98

And the dsitribution that satisfy this is X\sim Binom(n=9,p=0.75

Step-by-step explanation:

Previous concepts

The binomial distribution is a "DISCRETE probability distribution that summarizes the probability that a value will take one of two independent values under a given set of parameters. The assumptions for the binomial distribution are that there is only one outcome for each trial, each trial has the same probability of success, and each trial is mutually exclusive, or independent of each other".

Let X the random variable of interest, on this case we now that:

X \sim Binom(n=4, p=1-0.25=0.75)

The probability mass function for the Binomial distribution is given as:

P(X)=(nCx)(p)^x (1-p)^{n-x}

Where (nCx) means combinatory and it's given by this formula:

nCx=\frac{n!}{(n-x)! x!}

Part a

P(X=0)=(4C0)(0.75)^0 (1-0.75)^{4-0}=0.0039  

P(X=1)=(4C1)(0.75)^1 (1-0.75)^{4-1}=0.0469  

P(X=2)=(4C2)(0.75)^2 (1-0.75)^{4-2}=0.211  

P(X=3)=(4C3)(0.75)^2 (1-0.75)^{4-3}=0.422  

P(X=4)=(4C4)(0.75)^2 (1-0.75)^{4-4}=0.316

Part b

The expected value is givn by:

E(X) = np = 4*0.75=3

Part c

For the standard deviation we have this:

Sd(X) =\sqrt{np(1-p)}=\sqrt{4*0.75*(1-0.75)}=0.866

Part d

For this case the sample size needs to be higher or equal to 9. Since we need a value such that:

P(X \geq 3) \geq 0.98

And the dsitribution that satisfy this is X\sim Binom(n=9,p=0.75

We can verify this using the following code:

"=1-BINOM.DIST(3,9,0.75,TRUE)" and we got 0.99 and the condition is satisfied.

4 0
3 years ago
pipe A is a inlet pipe filling the tank at 8000l/hr and pipe B is the outlet pipe which empties the tank in 3 hours.the capacity
lawyer [7]
Well , in 3 hours , the amount of the tank that will be filled with pipe A = 8,000 x 3 = 24,000 L

Pipe could empty that 24,000 L in 3 hours.

So the capacity is  around 6 - 8 L .  .. . we need additional information to find out for sure

hope this helps
8 0
3 years ago
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