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denis23 [38]
2 years ago
5

In 2003, the population of Zimbabwe was about 12.6 million people, which is 1 million more than 4 times the population in 1950.

Write and solve an equation to find the approximate population p of Zimbabwe in 1950
Mathematics
1 answer:
alina1380 [7]2 years ago
8 0
Population of Zimbabwe in 2003 = 12.6 million
which is 1 million more than 4 times the population in 1950 
<span>approximate population p of Zimbabwe in 1950 = ?
12.6 million = </span>12,600,000
and 1 million = 1,000,000

The equation becomes:

12,600,000=1,000,000+4p

where p is the population in 1950

now solve for p,

12,600,000-1,000,000=4p

11,600,000=4p

11,600,000/4=p

p=2,900,000

which is the population in 1950

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The solution
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<span>s = 3s - 4 </span>
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<span>Check this: </span>

<span>m - 4 = 3s - 3 </span>
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5 0
1 year ago
Which shows the correct substitution of the values a, b, and c from the equation 0 = – 3x2 – 2x + 6 into the quadratic formula?
lbvjy [14]

Answer:

x = StartFraction negative

(negative 2) plus or minus StartRoot (negative 2) squared minus 4 (negative 3)(6) EndRoot Over 2(negative 3) EndFraction

Step-by-step explanation:

0 = – 3x2 – 2x + 6

It can still be written as

– 3x2 – 2x + 6 =0

Quadratic formula=

-b+or-√b^2-4ac/2a

Where

a=-3

b=-2

c=6

x= -(-2)+ or-√(-2)^2-4(-3)(6)/2(-3)

x = StartFraction negative

(negative 2) plus or minus StartRoot (negative 2) squared minus 4 (negative 3)(6) EndRoot Over 2(negative 3) EndFraction

8 0
1 year ago
The Fortune 500 is comprised of Fortune magazine's annual list of the finite population of the top 500 companies in the U.S. ran
Serjik [45]

Answer:

The standard error of the proportion is 0.0367.

Step-by-step explanation:

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean \mu and standard deviation \sigma, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean \mu and standard deviation s = \frac{\sigma}{\sqrt{n}}.

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

For a proportion p in a sample of size n, the standard error is s = \sqrt{\frac{p(1-p)}{n}}

In this question:

p = 0.16, n = 100

So

s = \sqrt{\frac{0.16*0.84}{100}} = 0.0367

The standard error of the proportion is 0.0367.

4 0
1 year ago
The probability of drawing two aces from a standard deck is 0.0059. We know this probability, but we don't know if the first car
ivanzaharov [21]

Answer:

Option C is right

C. They are independent because, based on the probability, the first ace was replaced before drawing the second ace.

Step-by-step explanation:

Given that the  probability of drawing two aces from a standard deck is 0.0059

If first card is drawn and replaced then this probability would change.  By making draws with replacement we make each event independent of the other

Drawing ace in I draw has probability equal to 4/52, when we replace the I card again drawing age has probability equal to same 4/52

So if the two draws are defined as event A and event B,  the events are  independent

C. They are independent because, based on the probability, the first ace was replaced before drawing the second ace.

4 0
1 year ago
What is the following simplified product? Assume x greater-than-or-equal-to 0 (StartRoot 10 x Superscript 4 Baseline EndRoot min
9966 [12]

Answer:

\bold{10x^4\sqrt{6}+x^3\sqrt{30x}-10x^4\sqrt{3}-x^3\sqrt{15x}}

Step-by-step explanation:

To find:

Simplified product of:

(\sqrt{10x^4}-x\sqrt{5x^2})(2\sqrt{15x^4}+\sqrt{3x^3})

Solution:

First of all, let us have a look at some of the formula:

1. (a+b) (c+d) = ac+bc+ad+bd

2. a^b\times a^c =a^{b+c }

3. \sqrt{a^{2b}} = \sqrt{a^b.a^b}=a^b

4. \sqrt a  \times \sqrt b = \sqrt{a\times b}

Now, let us apply the above formula to solve the given expression.

(\sqrt{10x^4}-x\sqrt{5x^2})(2\sqrt{15x^4}+\sqrt{3x^3})\\\\\Rightarrow(\sqrt{10x^4})(2\sqrt{15x^4})+(\sqrt{10x^4})(\sqrt{3x^3})-(x\sqrt{5x^2})(2\sqrt{15x^4})-(x\sqrt{5x^2})(\sqrt{3x^3})\\\\\Rightarrow2\sqrt{150x^8}+\sqrt{30x^7}-2x\sqrt{75x^6}-x\sqrt{15x^5}\\\\\Rightarrow\bold{10x^4\sqrt{6}+x^3\sqrt{30x}-10x^4\sqrt{3}-x^3\sqrt{15x}}

The answer is:

\bold{10x^4\sqrt{6}+x^3\sqrt{30x}-10x^4\sqrt{3}-x^3\sqrt{15x}}

8 0
2 years ago
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