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3 years ago

Cecilia ran 2 1/5 miles each day for one week. How many miles did she run altogether that week, including the weekend?

Mathematics

1 answer:

nordsb [41]3 years ago

8 0

Answer is 15.4 miles

Step-by-step explanation:

So she ran each day for a week and including the weekend is 7 days so if we do 7 times 2.20 it should equal 15.4 miles, also 2 1/5= 2.20 or 2.2 same thing.. Hope this helps, Also I hope I read the question right :/

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Evaluate Y=3x-5 when x=6

Harman [31]

7 + (-3x^2) for x = 0 7 + [-3(0)^2]; = 7 + -3(0); = 7 - 0; = 7

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3 years ago

Y(y + 4) - y 2 = 6 is a quadratic equation. True or False

Ugo [173]

Y(y + 4) - y2 = 6
y2 + 4y - y2 = 6
4y = 6 ⇒ False

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3 years ago

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After 7 years, Abner earned $1575 in simple interest from a CD into which he initially deposited $6000. What was the annual inte

Radda [10]

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Suppose a population grows according to the logistic equation but is subject to a constant total harvest rate of H. If N(t) is t

Elenna [48]

Answer:

a) Equilibrium point : [ 947, 53 ]

b) N = 947 is stable equilibrium, N = 53 is unstable equilibrium

c) N0, the population will not go extinct

Step-by-step explanation:

Given that;

r = 2, k = 1000, H = 100

dN/dT = R(1 - N/k)N - H

so we substitute

dN/dt = 2( 1 - N/1000)N - 100

now for equilibrium solution, dN/dt = 0

2( 1 - N/1000)N - 100 = 0

((1000 - N)/1000)N = 50

N^2 - 1000N + 50000 = 0

N = 1000 ± √(-1000)² - 4(1)(50000)) / 2(1)

N = 947.213 OR 52.786

approximately

N = 947 OR 53

Therefore Equilibrium point : [ 947, 53 ]

g(N) = 2( 1 - N/1000)N - 100

= 2N - N²/500 - 100

g'(N) = 2 - N/250

SO AT 947

g'(N) = g'(947) = 2 - 947/250 = -1.788 which is less than (<) 0

so N = 947 is stable equilibrium

now AT 53

g'(N) = g"(53) = 2 - 53/250 = 1.788 which is greater than (>) 0

so N = 53 is unstable equilibrium

The capacity k=1000

If the population is less than 53 then the population will become extinct but since the capacity is equal to 1000 then the population will not go extinct.

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3 years ago

Miranda has a bag of marbles with 5 blue marbles, 1 white marbles, and 1 red marbles. Find the following probabilities of Mirand

kvasek [131]

Answer:

(a) $\frac{5}{7},\frac{1}{6}$ (b) $\frac{1}{7},\frac{1}{6}$ (c) $\frac{5}{7},\frac{2}{3},\frac{3}{5}$

Step-by-step explanation:

GIVEN: Miranda has a bag of marbles with $5$ blue marbles, $1$ white marbles, and

TO FIND: a) A Blue, then a red Preview

,b)A red, then a white Preview

,c) A Blue, then a Blue, then a Blue.

SOLUTION:

Total marbles in bag $=7$

(a)

Probability of drawing blue marble $=\frac{\text{total blue marble}}{\text{total marbles}}$

$=\frac{5}{7}$

As marble is not returned to bag,

Probability of drawing red marble $=\frac{\text{total red marble}}{\text{total marbles}}$

$=\frac{1}{6}$

(b)

Probability of drawing red marble $=\frac{\text{total red marble}}{\text{total marbles}}$

$=\frac{1}{7}$

As marble is not returned to bag,

Probability of drawing white marble $=\frac{\text{total white marble}}{\text{total marbles}}$

$=\frac{1}{6}$

(c)

Probability of drawing blue marble $=\frac{\text{total blue marble}}{\text{total marbles}}$

$=\frac{5}{7}$

As marble is not returned to bag,

Probability of drawing second blue marble $=\frac{\text{total blue marble}}{\text{total marbles}}$

$=\frac{4}{6}=\frac{2}{3}$

As marble is not returned to the bag

Probability of drawing third blue marble $=\frac{\text{total blue marble}}{\text{total marbles}}$

$=\frac{3}{5}$

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3 years ago