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2 years ago

5

Match each equation to the situation it represents. Yin spends 10 hours on homework this week. She spends 5 hours of science hom

ework and then answers 35 math problems

1 answer:

cestrela7 [59]2 years ago

5 0

Answer:

35x+5=10

Step-by-step explanation:

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In 2002, the Centers for Disease Control and Prevention (CDC) reported that 8% of women married for the first time by their 18th

ehidna [41]

Question:

In 2002, the Centers for Disease Control and Prevention (CDC) reported that 8% of women married for the first time by their 18th birthday, 25% married by their 20th birthday, and 76% married by their 30th birthday. Based on these data, what is the probability that in a family with two daughters, the first and second daughter will be married by the following ages? (Enter your answers to four decimal places.) (a) 18 years of age (b) 20 years of age (c) 30 years of age

Answer:

A.) 0.0064

B.) 0.0625

C.) 0.5776

Step-by-step explanation:

Given the following :

Married by 18th birthday = 8% = 0.08

Married by 20th birthday = 25% = 0.25

Married by 30th birthday = 76% = 0.76

In a family with two(2) daughters :

P(First daughter will be married by 18) = 0.08

P(second daughter will be married by 18) = 0.08

P(1st and 2nd married by 18) = (0.08×0.08) = 0.0064

B.)

P(First daughter will be married by 20) = 0.25

P(second daughter will be married by 20) = 0.25

P(1st and 2nd married by 20) = (0.25×0.25) = 0.0625

C.)

P(First daughter will be married by 30) = 0.76

P(second daughter will be married by 30) = 0.76

P(1st and 2nd married by 30) = (0.76×0.76) = 0.5776

6 0

2 years ago

What is the percent of decrease from 40 to 26?

inessss [21]

Answer:

65%

Step-by-step explanation:

STEP ONE:

Find unit percent of 40. You should get 0.4.

STEP TWO:

Divide 26 by 0.4 and get 65.

So, the answer is 65%.

3 0

2 years ago

Read 2 more answers

Prove that the sum of all natural numbers is -1/12.

ankoles [38]

This is not true. The infinite series

$\displaystyle\sum_{n=1}^\infty n$

converges if and only if the sequence of its partial sums converges. The $k$ -th partial sum is

$\displaystyle\sum_{n=1}^kn=\frac{k(k+1)}2$

but clearly this diverges as $k$ gets arbitrarily large.

5 0

3 years ago

After an antibiotic tablet is taken, the concentration of the antibiotic in the bloodstream is modelled by the function C(t)=8(e

Alexxx [7]

Answer:

the maximum concentration of the antibiotic during the first 12 hours is 1.185 $\mu g/mL$ at t= 2 hours.

Step-by-step explanation:

We are given the following information:

After an antibiotic tablet is taken, the concentration of the antibiotic in the bloodstream is modeled by the function where the time t is measured in hours and C is measured in $\mu g/mL$

$C(t) = 8(e^{(-0.4t)}-e^{(-0.6t)})$

Thus, we are given the time interval [0,12] for t.

We can apply the first derivative test, to know the absolute maximum value because we have a closed interval for t.
The first derivative test focusing on a particular point. If the function switches or changes from increasing to decreasing at the point, then the function will achieve a highest value at that point.

First, we differentiate C(t) with respect to t, to get,

$\frac{d(C(t))}{dt} = 8(-0.4e^{(-0.4t)}+ 0.6e^{(-0.6t)})$

Equating the first derivative to zero, we get,

$\frac{d(C(t))}{dt} = 0\\\\8(-0.4e^{(-0.4t)}+ 0.6e^{(-0.6t)}) = 0$

Solving, we get,

$8(-0.4e^{(-0.4t)}+ 0.6e^{(-0.6t)}) = 0\\\displaystyle\frac{e^{-0.4}}{e^{-0.6}} = \frac{0.6}{0.4}\\\\e^{0.2t} = 1.5\\\\t = \frac{ln(1.5)}{0.2}\\\\t \approx 2$

At t = 0

$C(0) = 8(e^{(0)}-e^{(0)}) = 0$

At t = 2

$C(2) = 8(e^{(-0.8)}-e^{(-1.2)}) = 1.185$

At t = 12

$C(12) = 8(e^{(-4.8)}-e^{(-7.2)}) = 0.059$

Thus, the maximum concentration of the antibiotic during the first 12 hours is 1.185 $\mu g/mL$ at t= 2 hours.

4 0

2 years ago

I need plz help fast

Bad White [126]

Answer:

<h2>-4a²b³ (-6+ 7a³)</h2><h2 />

Step-by-step explanation:

24a²b³ - 28a⁵b³

apply exponential rule:

24a²b³ - 28a²a³b³

re-write 24 as 6*4 and -28 as 7*4

6*4a²b³ + 7*4a²a³b³

factor common terms

-4a²b³ (-6+ 7a³)

3 0

2 years ago