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andreev551 [17]

3 years ago

6

The African bush elaphent weighs between 4.4 tons and 7.7 tons What are it's least and greatest weighs , rounded to the nearest

ton

2 answers:

True [87]3 years ago

8 0

4.4=4 and 7.7=7.8 that's the answer

djyliett [7]3 years ago

6 0

The least would be 4 because 4.4 rounds to 4
The greatest would be 7.8 because 7.7 rounds to 7.8

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City A had a population of 10000 in the year 1990. City A’s population grows at a constant rate of 3% per year. City B has a pop

Georgia [21]

Answer:

City A and city B will have equal population 25years after 1990

Step-by-step explanation:

Given

Let

$t \to$ years after 1990

$A_t \to$ population function of city A

$B_t \to$ population function of city B

<u>City A</u>

$A_0 = 10000$ ---- initial population (1990)

$r_A =3\%$ --- rate

<u>City B</u>

$B_{10} = \frac{1}{2} * A_{10}$ ----- t = 10 in 2000

$A_{20} = B_{20} * (1 + 20\%)$ ---- t = 20 in 2010

Required

When they will have the same population

Both functions follow exponential function.

So, we have:

$A_t = A_0 * (1 + r_A)^t$

$B_t = B_0 * (1 + r_B)^t$

Calculate the population of city A in 2000 (t = 10)

$A_t = A_0 * (1 + r_A)^t$

$A_{10} = 10000 * (1 + 3\%)^{10}$

$A_{10} = 10000 * (1 + 0.03)^{10}$

$A_{10} = 10000 * (1.03)^{10}$

$A_{10} = 13439.16$

Calculate the population of city A in 2010 (t = 20)

$A_t = A_0 * (1 + r_A)^t$

$A_{20} = 10000 * (1 + 3\%)^{20}$

$A_{20} = 10000 * (1 + 0.03)^{20}$

$A_{20} = 10000 * (1.03)^{20}$

$A_{20} = 18061.11$

From the question, we have:

$B_{10} = \frac{1}{2} * A_{10}$ and $A_{20} = B_{20} * (1 + 20\%)$

$B_{10} = \frac{1}{2} * A_{10}$

$B_{10} = \frac{1}{2} * 13439.16$

$B_{10} = 6719.58$

$A_{20} = B_{20} * (1 + 20\%)$

$18061.11 = B_{20} * (1 + 20\%)$

$18061.11 = B_{20} * (1 + 0.20)$

$18061.11 = B_{20} * (1.20)$

Solve for B20

$B_{20} = \frac{18061.11}{1.20}$

$B_{20} = 15050.93$

$B_{10} = 6719.58$ and $B_{20} = 15050.93$ can be used to determine the function of city B

$B_t = B_0 * (1 + r_B)^t$

For: $B_{10} = 6719.58$

We have:

$B_{10} = B_0 * (1 + r_B)^{10}$

$B_0 * (1 + r_B)^{10} = 6719.58$

For: $B_{20} = 15050.93$

We have:

$B_{20} = B_0 * (1 + r_B)^{20}$

$B_0 * (1 + r_B)^{20} = 15050.93$

Divide $B_0 * (1 + r_B)^{20} = 15050.93$ by $B_0 * (1 + r_B)^{10} = 6719.58$

$\frac{B_0 * (1 + r_B)^{20}}{B_0 * (1 + r_B)^{10}} = \frac{15050.93}{6719.58}$

$\frac{(1 + r_B)^{20}}{(1 + r_B)^{10}} = 2.2399$

Apply law of indices

$(1 + r_B)^{20-10} = 2.2399$

$(1 + r_B)^{10} = 2.2399$ --- (1)

Take 10th root of both sides

$1 + r_B = \sqrt[10]{2.2399}$

$1 + r_B = 1.08$

Subtract 1 from both sides

$r_B = 0.08$

To calculate $B_0$ , we have:

$B_0 * (1 + r_B)^{10} = 6719.58$

Recall that: $(1 + r_B)^{10} = 2.2399$

So:

$B_0 * 2.2399 = 6719.58$

$B_0 = \frac{6719.58}{2.2399}$

$B_0 = 3000$

Hence:

$B_t = B_0 * (1 + r_B)^t$

$B_t = 3000 * (1 + 0.08)^t$

$B_t = 3000 * (1.08)^t$

The question requires that we solve for t when:

$A_t = B_t$

Where:

$A_t = A_0 * (1 + r_A)^t$

$A_t = 10000 * (1 + 3\%)^t$

$A_t = 10000 * (1 + 0.03)^t$

$A_t = 10000 * (1.03)^t$

and

$B_t = 3000 * (1.08)^t$

$A_t = B_t$ becomes

$10000 * (1.03)^t = 3000 * (1.08)^t$

Divide both sides by 10000

$(1.03)^t = 0.3 * (1.08)^t$

Divide both sides by $(1.08)^t$

$(\frac{1.03}{1.08})^t = 0.3$

$(0.9537)^t = 0.3$

Take natural logarithm of both sides

$\ln(0.9537)^t = \ln(0.3)$

Rewrite as:

$t\cdot\ln(0.9537) = \ln(0.3)$

Solve for t

$t = \frac{\ln(0.3)}{ln(0.9537)}$

$t = 25.397$

Approximate

$t = 25$

7 0

3 years ago

suppose andrew slept for eight hours one night. Ten percent of his sleep hours were before mignight. What.time did andrew go to

m_a_m_a [10]

Answer: 11:12 PM.
explanation:
- 10% of 8 is 0.8. (10% is 0.1, and you multiply this by 8)
- multiply 0.8 by 60 so you can get the number of minutes before midnight (this is 48).
- 48 minutes before midnight is 11:12.

6 0

3 years ago

A four year old is going to spin around with his arms stretched out 100 times. From past experience, his father knows it takes a

Sati [7]

Answer:

$P(X \geq0.55) \leq 0.22$

Step-by-step explanation:

Using central Limit Theorem (CLT), The sum of 100 random variables;

$Y=X_1+X_2+...+X_{100}$ is approximately normally distributed with

Y ~ N (100 × $\frac{1}{3^2}$ ) = N ( 50, $\frac{100}{9}$ )

The approximate probability that it will take this child over 55 seconds to complete spinning can be determined as follows;

N ( 50, $\frac{100}{9}$ )

$P(Y>55) =P(Z>\frac{55-50}{10/3})$

$P(Y>55) =P(Z>1.5)$

$P(Y>55) =\phi (-1.5)$

$P(Y>55) =0.0668$

Using Chebyshev's inequality:

$P(|X-\mu\geq K)\leq \frac{\sigma^2}{K^2}$

Let assume that X has a symmetric distribution:

Then:

$2P(X-\mu\geq K)\leq) \frac{\sigma^2}{K^2}$

$2P(X \geq \mu+K)\leq) \frac{\sigma^2}{K^2}$

$2P(X\geq0.5+0.05)\leq \frac{\frac{1}{\frac{3^2}{100} } }{0.05^2}$ where: ( $\sigma^2 = \frac{1}{3^2/100}$ )

$P(X \geq0.55) \leq 0.22$

6 0

3 years ago

Order these numbers from least to greatest. 9.4061 , 9.07 , 9.007 , 9.4

klio [65]

9.4061,
9.07,
9.007,
9.4
Because ones are equal for all numbers, we are going to begin to compare from the second digits (tenths).
Greater :
9.4061 and 9.4.
Greatest is this group is 9.4061, then will go 9.4.

Smaller :
9.07 and 9.007
Greatest in this group is 9.07, and then will go 9.007.

So, from greatest to smallest : 9.4061 , 9.4 , 9.07 , 9.007.

From least to greatest :
9.007, 9.07, 9.4, 9.4061.

3 0

3 years ago

find the mean, median, and mode of the data set round to the nearest tenth 15, 1 , 4, 4, 8, 7, 15, 4, 5

Dmitriy789 [7]

$15,\ 1,\ 4,\ 4,\ 8,\ 7,\ 15,\ 4,\ 15,\ 4,\ 5\to\underbrace{1,\ 4,\ 4,\ 4,\ 4,\ 5,\ 7,\ 8,\ 15,\ 15,\ 15}_{11}\\\\mean=\dfrac{1+4+4+4+4+5+7+8+15+15+15}{11}=\dfrac{82}{11}\approx7.5\\\\median:1,\ 4,\ 4,\ 4,\ 4,\ \boxed{5},\ 7,\ 8,\ 15,\ 15,\ 15\\\\mode:1,\ \underbrace{4,\ 4,\ 4,\ 4}_{4},\ 5,\ 7,\ 8,\ 15,\ 15,\ 15$

<h3>Answer: mean = 7.5, median = 5, mode = 4</h3>

6 0

3 years ago

Read 2 more answers