Ask question

Ask question

All categories

3 years ago

15

The average adult has 1.2 to 1.5 gallons of blood in their body how much blood does the average adult have in liters round your

answer to the nearest tenth of aliter

1 answer:

Lisa [10]3 years ago

3 0

Answer:4.5 to 5.7 L

Step-by-step explanation:

You might be interested in

7/6 - (1/4 + 1/3) =??

Hitman42 [59]

Answer:

7/12

Step-by-step explanation:

7/6 - (1/4 + 1/3) =

7/6 - 1/4 - 1/3 = 14/12 - 3/12 - 4/12 = 7/12

7 0

3 years ago

Read 2 more answers

During a football game a concession stand sold a family three hamburgers and two hotdogs for a total of $13 it's sold another fa

guajiro [1.7K]

Answer:

The price of each hamburger is $3

The price of each hot dogs is $2 .

Step-by-step explanation:

Given as :

The total price of 3 hamburger and 2 hot dogs = $13

The total price of 2 hamburger and 5 hot dogs = $16

Let The price of each hamburger = $x

Let The price of each hot dogs = $y

<u>Now, According to question</u>

3 x + 2 y = 13 .........A

2 x + 5 y = 16 .......B

Now, Solving to eq A and B

3 × (2 x + 5 y ) - 2 × (3 x + 2 y ) = 3 × 16 - 2 × 13

Or, (6 x + 15 y) - (6 x + 4 y) = 48 - 26

Or, (6 x - 6 x) + (15 y - 4 y) = 22

Or, 0 + 11 y = 22

∴ y = $\dfrac{22}{11}$

i.e y = $2

so, The price of each hot dogs = y = $2

<u>Now, Put the value of y into eq B</u>

i.e 2 x + 5 y = 16

or, 2 x + 5 × 2 = 16

or, 2 x = 16 - 10

or, 2 x = 6

∴ x = $\dfrac{6}{2}$

i.e x = $3

So, The price of each hamburger = x = $3

Hence, The price of each hamburger is $3 and The price of each hot dogs is $2 . Answer

3 0

3 years ago

Cassie rolls a fair number cube with 6 faces labeled 1 through 6 she rolls the number cube 300 times which results is most likel

ollegr [7]

It depend upon what you roll it could be 6 because there are 6 faces are the dice

3 0

3 years ago

Using f(x) = 8x + 5 and g(x) = 7x - 2, find:g(f(6))

iren2701 [21]

Answer:

369

Step-by-step explanation:

g(f(x)) simply means to put the whole expression of function f(x) into the place of "x" in g(x).

and to find the functional value for a specific x we just need to calculate functional value for f(x) and use that result as input for g(x).

f(6) = 8×6 + 5 = 48 + 5 = 53

and now

g(53) = 7×53 - 2 = 371 - 2 = 369

to check we do the general functional substitution :

g(f(x)) = 7×(8x+5) - 2 = 56x + 35 - 2 = 56x + 33

g(f(6)) = 56×6 + 33 = 336 + 33 = 369

correct

5 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