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3 years ago

10

Luis made some trail mix. he mixed 4 and 2/3 cups of popcorn, 1 and 1/4 cups of peanuts, 1 and 1/3 of raisins, and 3/4 cup of su

nflower seeds. he gave 5 of his friends an equal amount of trail mix each. how much did each friend get?

1 answer:

vodka [1.7K]3 years ago

3 0

Add 4and 2/3 cups, 1 and 1/4 cups,1 and 1/3 cups and 3/4 cups all together and divide it by 5

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When Akiko measured a rose, its height was 5.8 in. After 10 weeks, the height was 1 1/3 times the original height. What was the

kramer

The solution is in the attachment

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3 years ago

Express the probability that a die wil land on a number less than 5 as a fraction

Rashid [163]

Answer:

$\frac{2}{3}$

Step-by-step explanation:

Values less than 5 on a die are 1, 2, 3, 4 ← 4 values out of a possible 6

P( < 5) = $\frac{4}{6}$ = $\frac{2}{3}$ in simplest form

7 0

3 years ago

What value(s) of x will make each equation below true

Dahasolnce [82]

We need to solve for x. Let's try problem b:

$2x-6=3x+1-x-7$

Let us first combine line terms. 3x and -x as well as 1 and -7 can be combined. Let's do that:

$2x-6=2x-6$

Since this is true, your answer would be:

All real numbers

------------------------------------------------------------------

Let's solve for problem c:

$3x+1=43$

Let's isolate x, so subtract 1 from both sides:

$3x=42$

Since x can't have a coefficient, divide both sides by 3:

$\frac{3x}{3}=\frac{42}{3}$

$x=14$

So, only the value of 14 would make this equation true.

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Let's try problem d:

$4x-1=4x+7$

Let's get our whole numbers on the right side. Add 1 to both sides:

$4x=4x+8$

Subtract 4x from the right side on both sides:

$0=8$

Since this is not true, your answer would be:

No solution

7 0

3 years ago

Uninhibited growth can be modeled by exponential functions other than A(t) =Upper A 0 e Superscript kt. For example, if an i

laila [671]

The question is incomplete. Here is the complete question.

Uninhibited growth can be modeled by exponential functions other than $A(t)=A_{0}e^{kt}$ . for example, if an initial population P₀ requires n units of time to triple, then the function $P(t)=P_{0}(3)^{\frac{t}{n} }$ models the size of the population at time t. An insect population grows exponentially. Complete the parts a through d below.

a) If the population triples in 30 days, and 50 insects are present initially, write an exponential function of the form $P(t)=P_{0}(3)^{\frac{t}{n} }$ that models the population.

b) What will the population be in 47 days?

c) When wil the population reach 750?

d) Express the model from part (a) in the form $A(t)=A_{0}e^{kt}$ .

Answer: a) $P(t)=50(3)^{\frac{t}{30} }$

b) P(t) = 280 insects

c) t = 74 days

d) $A(t)=50e^{0.037t}$

Step-by-step explanation:

a) n is time necessary to triple the population of insects, i.e., n = 30 and P₀ = 50. So, Exponential equation for growth is

$P(t)=50(3)^{\frac{t}{30} }$

b) In t = 47 days:

$P(t)=50(3)^{\frac{t}{30} }$

$P(47)=50(3)^{\frac{47}{30} }$

$P(47)=50(3)^{1.567}$

P(47) = 280

In 47 days, population of insects will be 280

c) P(t) = 750

$750=50(3)^{\frac{t}{30} }$

$\frac{750}{50}=(3)^{\frac{t}{30} }$

$(3)^{\frac{t}{n} }=15$

Using the property <u>Power</u> <u>Rule</u> of logarithm:

$log(3)^{\frac{t}{30} }=log15$

$\frac{t}{30}log(3)=log15$

$t=\frac{log15}{log3} .30$

t = 74

To reach a population of 750 insects, it will take 74 days

d) To express the population growth into the described form, determine the constant k, using the following:

A(t) = 3A₀ and t = 30

$A(t)=A_{0}e^{kt}$

$3A_{0}=A_{0}e^{30k}$

$3=e^{30k}$

Use Power Rule again:

$ln3=ln(e^{30k})$

$ln3=30k$

$k=\frac{ln3}{30}$

k = 0.037

Equation for exponential growth will be:

$A(t)=50e^{0.037t}$

3 0

3 years ago

I need help on this one

Tanzania [10]

What is it? I will help. :)

4 0

3 years ago