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2 years ago

Help please asappppppppp

Mathematics

1 answer:

Aleonysh [2.5K]2 years ago

3 0

Answer:

First Equation: 6a-23

Second Equation: 5a-20

Step-by-step explanation:

So, I think you want me to answer part 2...

For the first equation, we can see that the denominator stays the exact same, meaning the numerator will also not change.

For the second equation, we can see the denominator changed from a-3 to 5a-15. The denominator was simply multiplied by 5. Since the denominator was multiplied by five we must multiply the numerator by five as well.

5(a-4)

5a-20

Hope this helps :)

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You deposit $300 in a savings account. The account earns 2% simple interest per year.

Sholpan [36]

Answer:

Answer:

a. ----> $10

b. ----> $110

Step-by-step explanation:

principle ( p ) = $100

time ( t ) = 10 years

rate ( r ) = 11%

simple interest = (p × r × t)÷ 100

= ( $100 × 11 × 10 )÷ 100

= 11000 ÷ 100

= $110

interest = simple interest - principle

interest = $110 - $100

= $10

8 0

2 years ago

Need help!! ASAP!! Geometry

kodGreya [7K]

Total 52 students.

52 - 17 = 35 students left
35 - 5 = 30 students left
30 - 2 = 28 students left

Each class has 11 students that play ONE sport, which means 22 total in 2 classes. You're on your own now! It's not that hard, I'm sure you'll figure it out.

7 0

3 years ago

At a specific point on a highway, vehicles arrive according to a Poisson process. Vehicles are counted in 12 second intervals, a

morpeh [17]

Answer: a) 4.6798, and b) 19.8%.

Step-by-step explanation:

Since we have given that

P(n) = $\dfrac{15}{120}=0.125$

As we know the poisson process, we get that

$P(n)=\dfrac{(\lambda t)^n\times e^{-\lambda t}}{n!}\\\\P(n=0)=0.125=\dfrac{(\lambda \times 14)^0\times e^{-14\lambda}}{0!}\\\\0.125=e^{-14\lambda}\\\\\ln 0.125=-14\lambda\\\\-2.079=-14\lambda\\\\\lambda=\dfrac{2.079}{14}\\\\0.1485=\lambda$

So, for exactly one car would be

P(n=1) is given by

$=\dfrac{(0.1485\times 14)^1\times e^{-0.1485\times 14}}{1!}\\\\=0.2599$

Hence, our required probability is 0.2599.

a. Approximate the number of these intervals in which exactly one car arrives

Number of these intervals in which exactly one car arrives is given by

$0.2599\times 18=4.6798$

We will find the traffic flow q such that

$P(0)=e^{\frac{-qt}{3600}}\\\\0.125=e^{\frac{-18q}{3600}}\\\\0.125=e^{-0.005q}\\\\\ln 0.125=-0.005q\\\\-2.079=-0.005q\\\\q=\dfrac{-2.079}{-0.005}=415.88\ veh/hr$

b. Estimate the percentage of time headways that will be 14 seconds or greater.

so, it becomes,

$P(h\geq 14)=e^{\frac{-qt}{3600}}\\\\P(h\geq 14)=e^{\frac{-415.88\times 14}{3600}}\\\\P(h\geq 14)=0.198\\\\P(h\geq 14)=19.8\%$

Hence, a) 4.6798, and b) 19.8%.

7 0

3 years ago

stat: sample of 780 adults in the U.S.A., it was found that 80 of those had a pinworm infestation. You want to find the 99% conf

Anettt [7]

Answer: 554

Step-by-step explanation:

If prior population proportion is known, then the formula to find the sample size is given by :-

$n=p(1-p)(\dfrac{z_{\alpha/2}}{E})^2$

As per given description, we have

p= 0.1

E=0.025

Critical z-value for 95% confidence : $z_{\alpha/2}=1.96$

Then,

$n=0.1(1-0.1)(\dfrac{1.96}{0.025})^2=553.1904\approx554$

Hence, the minimum sample size required = 554.

7 0

3 years ago

Use the mapping shown above to show the value of the function f(x) at each point.

julsineya [31]

Answer:

The value of the function f(x) at each point is f(-1) = 5, f(0) = 3, f(1) = 5, f(2) = 11, f(3) = 21 ⇒ A

Step-by-step explanation:

Let us use the mapping shown to solve the question

∵ f(x) = y

∵ x is the domain

∵ y is the range

→ From the figure use x from the domain and y from the range, where

each arrow pointed at the corresponding value y of x

∵ x = -1 and the corresponding value of y is 5

∴ f(-1) = 5

∵ x = 0 and the corresponding value of y is 3

∴ f(0) = 3

∵ x = 1 and the corresponding value of y is 5

∴ f(1) = 5

∵ x = 2 and the corresponding value of y is 11

∴ f(2) = 11

∵ x = 3 and the corresponding value of y is 21

∴ f(3) = 21

The value of the function f(x) at each point is f(-1) = 5, f(0) = 3, f(1) = 5, f(2) = 11, f(3) = 21

7 0

2 years ago

Help please asappppppppp​

Help please asappppppppp