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3 years ago

(01.07)

Mathematics

2 answers:

kotykmax [81]3 years ago

8 0

<span>Distributive Property</span>

sweet-ann [11.9K]3 years ago

5 0

I just took the test, the answer A Distributive Property is correct

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dalvyx [7]

Answer:

Step-by-step explanation:

10+8-3+5-12=

5 0

3 years ago

Pleasantburg has a population growth model of P(t)=at2+bt+P0 where P0 is the initial population. Suppose that the future populat

yulyashka [42]

Answer:

The population will reach 34,200 in February of 2146.

Step-by-step explanation:

Population in t years after 2012 is given by:

$P(t) = 0.8t^{2} + 6t + 19000$

In what month and year will the population reach 34,200?

We have to find t for which P(t) = 34200. So

$P(t) = 0.8t^{2} + 6t + 19000$

$0.8t^{2} + 6t + 19000 = 34200$

$0.8t^{2} + 6t - 15200 = 0$

Solving a quadratic equation:

Given a second order polynomial expressed by the following equation:

$ax^{2} + bx + c, a\neq0$ .

This polynomial has roots $x_{1}, x_{2}$ such that $ax^{2} + bx + c = a(x - x_{1})*(x - x_{2})$ , given by the following formulas:

$x_{1} = \frac{-b + \sqrt{\bigtriangleup}}{2*a}$

$x_{2} = \frac{-b - \sqrt{\bigtriangleup}}{2*a}$

$\bigtriangleup = b^{2} - 4ac$

In this question:

$0.8t^{2} + 6t - 15200 = 0$

So $a = 0.8, b = 6, c = -15200$

Then

$\bigtriangleup = 6^{2} - 4*0.8*(-15100) = 48356$

$t_{1} = \frac{-6 + \sqrt{48356}}{2*0.8} = 134.14$

$t_{2} = \frac{-6 - \sqrt{48356}}{2*0.8} = -141.64$

We only take the positive value.

134 years after 2012.

.14 of an year is 0.14*365 = 51.1. The 51st day of a year happens in February.

So the population will reach 34,200 in February of 2146.

6 0

3 years ago

? can guys help me please?

weqwewe [10]

Answer:

The range of this set of numbers is 9

Step-by-step explanation:

The range is the greatest number subtracted by the lowest number

in this problem that is 21-12

7 0

3 years ago

Read 2 more answers

The batteries from a certain manufacturer have a mean lifetime of 850 hours, with a standard deviation of 70 hours, assuming tha

matrenka [14]

To answer (a), we will need to find the Z value for 710 hours and 990 hours.

$\begin{gathered} For\text{ 710} \\ Z\text{ = }\frac{x-\mu}{\sigma} \\ \text{ = }\frac{710-850}{70} \\ \text{ = -2} \\ p\text{ =0.0228 } \\ For\text{ 990:} \\ Z\text{ = }\frac{990-850}{70} \\ \text{ =2} \\ p\text{ = 0.9772} \end{gathered}$ $\begin{gathered} To\text{ find P \lparen–2 \le Z \le 2\rparen} \\ =\text{ 0.9772 -0.0228} \\ =\text{ 0.9544} \\ =95.44\% \end{gathered}$

B) 68% of data lies between one standard deviation of the mean.

850 +70 = 920

850 - 70 = 780

5 0

2 years ago

Write an equation that reflects this relationship using x to represent the amount of fish in grams and y to represent the number

NeTakaya

Answer:

send the graph and ill solve it

Step-by-step explanation:

7 0

3 years ago