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3 years ago

Can you guys check this???

Mathematics

1 answer:

snow_tiger [21]3 years ago

6 0

On 4 and 6 because you are multiplying on both sides you need to flip the inequality signs
4 would be “e> -2”
6 would be “c<4”

Everything else was right
Hope this helps :D

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Work out m and c for the line: y = 2x - 3

saveliy_v [14]

Answer:

m = 2

c = -3

Explanation:

Equation of Line ⇒ y = mx +c

y = 2x - 3

m = 2

c = -3

3 0

3 years ago

Read 2 more answers

Please help if you could. f/2 + 9 = 4 -9 -9 f/2 =

dmitriy555 [2]

Answer:

$\frac{f}{2} +9=4$

Subtract 9 from both sides:

$\frac{f}{2}+9-9=4-9$

$\frac{f}{2} =-5$

Multiply both sides by 2:

$2*\frac{f}{2} =-5*2$

$f=-10$

8 0

3 years ago

A sample of a radioactive isotope had an initial mass of 360 mg in the year 1998 and

RUDIKE [14]

Answer:

193 mg

Step-by-step explanation:

Exponential decay formula:

$A_t = A_0e^r^t$
where Aₜ = mass at time t, A₀ = mass at time 0, r = decay constant (rate), t = time

Our known variables are:

1998 to the year 2004 is a total of t = 6 years.
The sample of radioactive isotope has an initial mass of A₀ = 360 mg at time 0 and a mass of Aₜ = 270 mg at time t.

Let's solve for the decay constant of this sample.

$270=360e^-^r^(^6^)$
$270=360e^-^6^r$
$\frac{3}{4} =e^-^6^r$
$\text{ln} (\frac{3}{4} )= \text{ln}(e^-^6^r)$
$\text{ln} (\frac{3}{4} )=-6r$
$r=-\frac{\text{ln}\frac{3}{4} }{6}$
$r=0.04794701$

Using our new variables, we can now solve for Aₜ at t = 7 years, since we go from 2004 to 2011.

Our new initial mass is A₀ = 270 mg. We solved for the decay constant, r = 0.04794701.

$A_t=270e^-^(^0^.^0^4^7^9^4^8^0^1^)^(^7^)$
$A_t=270e^-^0^.^3^3^5^6^2^9^0^7$
$A_t=193.01982213$

The expected mass of the sample in the year 2011 would be 193 mg.

3 0

3 years ago

Fill in the blanks to complete the following statements. Bold left parenthesis a right parenthesis For the shape of the distribu

Nana76 [90]

Answer:

For the shape of the distribution of the sample proportion to be approximately normal, it is required that np (1 -p ) greater than or equals 10.

Suppose the proportion of a population that has a certain characteristic is 0.35. The mean of the sampling distribution of ModifyingAbove p with caret from this population is mu Subscript ModifyingAbove p with caretequals0.35.

Step-by-step explanation:

Normal distribution is the shape data takes as a symmetrical bell shaped curve. Normal approximation can only be taken when np or np(1-p) greater than 10.

Fill in the blanks to complete the following statements:

For the shape of the distribution of the sample proportion to be approximately normal, it is required that np(1 - p)greater than or equals__10____.
Suppose the proportion of a population that has a certain characteristic is 0.35. The mean of the sampling distribution of ModifyingAbove p with caret from this population is mu Subscript ModifyingAbove p with caretequals__0.35____.

6 0

4 years ago

The unit rate of change of y with respect to x is the amount y changes for a change of one unit in x. Is the unit rate of change

Nataly [62]

Answer:

Option (a). The equation

Step-by-step explanation:

Rate of change of any equation or graph is represented by its slope 'm'.

In an equation y = mx + b

'm' represents the slope.

Given equation is,

y = 0.25x

Slope of the line m₁ = 0.25

From the graph attached,

A line passes through two points (0, 0) and (5, 1)

Slope of the line = $\frac{y_2-y_1}{x_2-x_1}$

m₂ = $\frac{1-0}{5-0}$

m₂ = 0.2

m₁ > m₂

Therefore, unit rate of change in y is greater in the given equation.

Option (a) will be the answer.

6 0

3 years ago

Read 2 more answers