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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
Mathematics
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
You might be interested in
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
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