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3 years ago

How to use a number line to solve 235+123

1 answer:

7 0

Plot the point 123 on the line and go to the right 235 units.
because there is the big number, just draw the line begin with 123 to 400.
each unit might increase by 10.
123, 133, 143, etc
you can increase bigger if you want

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Three cowboys share 25.6 pounds of bacon.how many bacon dose each cowboy receive?

ANTONII [103]

25.6 divided by 3 = 8.53

5 0

3 years ago

Find a solution to the linear equation 13x+y=26

Anna11 [10]

Answer:

y = -13x+26

Step-by-step explanation:

13x+y=26

y= -13x+26

8 0

3 years ago

Which inequality represents all possible solutions of −6n<−12?

muminat

Answer:

B n>2

Step-by-step explanation:

−6n<−12

Divide each side by -6, remembering to flip the inequality

-6n/-6 > -12/-6

n >2

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3 years ago

Animal populations are not capable of unrestricted growth because of limited habitat and food supplies. Under such conditions th

trasher [3.6K]

Answer:

(a) 100 fishes

(b) t = 10: 483 fishes

t = 20: 999 fishes

t = 30: 1168 fishes

(c)

$P(\infty) = 1200$

Step-by-step explanation:

Given

$P(t) =\frac{d}{1+ke^-{ct}}$

$d = 1200\\k = 11\\c=0.2$

Solving (a): Fishes at t = 0

This gives:

$P(0) =\frac{1200}{1+11*e^-{0.2*0}}$

$P(0) =\frac{1200}{1+11*e^-{0}}$

$P(0) =\frac{1200}{1+11*1}$

$P(0) =\frac{1200}{1+11}$

$P(0) =\frac{1200}{12}$

$P(0) = 100$

Solving (a): Fishes at t = 10, 20, 30

$t = 10$

$P(10) =\frac{1200}{1+11*e^-{0.2*10}} =\frac{1200}{1+11*e^-{2}}\\\\P(10) =\frac{1200}{1+11*0.135}=\frac{1200}{2.485}\\\\P(10) =483$

$t = 20$

$P(20) =\frac{1200}{1+11*e^-{0.2*20}} =\frac{1200}{1+11*e^-{4}}\\\\P(20) =\frac{1200}{1+11*0.0183}=\frac{1200}{1.2013}\\\\P(20) =999$

$t = 30$

$P(30) =\frac{1200}{1+11*e^-{0.2*30}} =\frac{1200}{1+11*e^-{6}}\\\\P(30) =\frac{1200}{1+11*0.00247}=\frac{1200}{1.0273}\\\\P(30) =1168$

Solving (c): $\lim_{t \to \infty} P(t)$

In (b) above.

Notice that as t increases from 10 to 20 to 30, the values of $e^{-ct}$ decreases

This implies that:

${t \to \infty} = {e^{-ct} \to 0}$

So:

The value of P(t) for large values is:

$P(\infty) = \frac{1200}{1 + 11 * 0}$

$P(\infty) = \frac{1200}{1 + 0}$

$P(\infty) = \frac{1200}{1}$

$P(\infty) = 1200$

5 0

3 years ago

Resistors are labeled 100 Ω. In fact, the actual resistances are uniformly distributed on the interval (95, 103). Find the mean

Zinaida [17]

Answer:

$E[R]$ = 99 Ω

$\sigma_R$ = 2.3094 Ω

P(98<R<102) = 0.5696

Step-by-step explanation:

The mean resistance is the average of edge values of interval.

Hence,

The mean resistance, $E[R] = \frac{a+b}{2} = \frac{95+103}{2} = \frac{198}{2}$ = 99 Ω

To find the standard deviation of resistance, we need to find variance first.

$V(R) = \frac{(b-a)^2}{12} =\frac{(103-95)^2}{12} = 5.333$

Hence,

The standard deviation of resistance, $\sigma_R = \sqrt{V(R)} = \sqrt5.333$ = 2.3094 Ω

To calculate the probability that resistance is between 98 Ω and 102 Ω, we need to find Normal Distributions.

$z_1 = \frac{102-99}{2.3094} = 1.299$

$z_2 = \frac{98-99}{2.3094} = -0.433$

From the Z-table, P(98<R<102) = 0.9032 - 0.3336 = 0.5696

5 0

4 years ago