All categories

3 years ago

In a fishing competition Johannes and Peter together caught 48 fish.

Mathematics

1 answer:

Alenkasestr [34]3 years ago

4 0

John caught 18 fishes and Peter caught 30 fishes.

Step-by-step explanation:

Let,

Fishes caught by John = x

Fishes caught by peter = y

Total fishes caught = 48

According to given statement;

x+y=48 Eqn 1

x= $\frac{3}{5}y$ Eqn 2

Putting Eqn 2 in Eqn 1;

$\frac{3}{5}y+y=48$

Multiplying each term by 5;

$5*\frac{3}{5}y+5*y=48*5\\3y+5y=240\\8y=240$

Dividing both sides by 8

$\frac{8y}{8}=\frac{240}{8}\\y=30$

Putting y=30 in Eqn 1

$x+30=48\\x=48-30\\x=18$

John caught 18 fishes and Peter caught 30 fishes.

Keywords: linear equation, substitution method

Learn more about substitution method at:

#LearnwithBrainly

You might be interested in

What is 2.40 before a 30 percent discount

artcher [175]

2.40 before a 30% discount is 0.72

7 0

3 years ago

Read 2 more answers

The rate of change of the number of mountain lions N(t) in a population is directly proportional to 725 - N(t), where t is the t

wolverine [178]

Answer:

a) The implied differential equation is $\frac{dN}{725 - N} = kdt$

b) The general equation is $N = 725 - C e^{-kt}$

c) The particular equation is $N = 725 - 325 e^{-0.49t}$

d) The population when t = 5, N(5) = 697 = 700( to the nearest 50)

Step-by-step explanation:

The rate of change of N(t) can be written as dN/dt

According to the question, $\frac{dN}{dt} \alpha (725 - N(t))$

$\frac{dN}{dt} = k (725 - N)\\\frac{dN}{725 - N} = kdt$

Integrating both sides of the equation

$\int {\frac{1 }{725 - N}} \, dN = \int {k} \, dt\\- ln (725 - N) = kt + C\\ ln (725 - N) = -(kt + C)\\725 - N = e^{-(kt + C)} \\725 - N = e^{-kt} * e^{-C} \\725 - N = C e^{-kt}\\N = 725 - C e^{-kt}$

When t = 0, N = 400

$400 = 725 - C e^{-k*0}\\400 = 725 - C\\C = 725 - 400\\C = 325$

When t = 3, N = 650

$650 = 725 - (325 * e^{-3k})\\325 * e^{-3k} = 75\\e^{-3k} = 75/325\\e^{-3k} = 0.23\\-3k = ln 0.23\\-3k = -1.47\\k = 1.47/3\\k = 0.49$

The equation for the population becomes:

$N = 725 - 325 e^{-0.49t}$

At t = 5, the population becomes:

$N = 725 - 325 e^{-0.49*5}\\N = 725 - 325 e^{-2.45}\\N = 696.95\\N(5) = 697$

N(5) = 700 ( to the nearest 50)

6 0

3 years ago

Summary

ziro4ka [17]

Answer:

what do u mean, never mind answer is square units

6 0

3 years ago

A cable company charges a flat fee of $50 per month plus $8 per month for each movie channel, m, and $4 per month for each sport

PSYCHO15rus [73]

T = 12( 50 + 8m + 4s)

T = 600 + 96m + 48s

7 0

3 years ago

Read 2 more answers

Suppose a researcher is testing the hypothesis Upper H 0: pequals0.4 versus Upper H 1: pless than0.4 and she finds the P-valu

andre [41]

Answer:

D. If the P-value for a particular test statistic is 0.33, she expects results at least as extreme as the test statistic in exactly 33 of 100 samples if the null hypothesis is true.

D. Since this event is not unusual, she will not reject the null hypothesis.

Step-by-step explanation:

Hello!

You have the following hypothesis:

H₀: ρ = 0.4

H₁: ρ < 0.4

Calculated p-value: 0.33

Remember: The p-value is defined as the probability corresponding to the calculated statistic if possible under the null hypothesis (i.e. the probability of obtaining a value as extreme as the value of the statistic under the null hypothesis).

In this case, you have a 33% chance of getting a value as extreme as the statistic value if the null hypothesis is true. In other words, you would expect results as extreme as the calculated statistic in 33 about 100 samples if the null hypothesis is true.

You didn't exactly specify a level of significance for the test, so, I'll use the most common one to make a decision: α: 0.05

Remember:

If p-value ≤ α, then you reject the null hypothesis.

If p-value > α, then you do not reject the null hypothesis.

Since 0.33 > 0.05 then I'll support the null hypothesis.

I hope it helps!

3 0

3 years ago