Points A, B, and C lie on the same line.

Cassidy wrote the linear equation y = 5 x + 4. Then, she wrote the equation of the line that is parallel to y = 5 x + 4 and that

Kryger [21]

Answer:

b=−5x+y

Step-by-step explanation:

Let's solve for b.

y=5x+b

Step 1: Flip the equation.

b+5x=y

Step 2: Add -5x to both sides.

b+5x+−5x=y+−5x

b=−5x+y

3 0

3 years ago

write the equation of a line that passes through (4,10) and is parallel to the line with the equation y=-3x+7

Alex_Xolod [135]

Y-20=1/3(x-4)

y = 1/3x+56/3

8 0

4 years ago

6/9=x/6 solve for the variable x

sammy [17]

$\frac{6}{9} = \frac{x}{6} \Leftrightarrow x = \frac{ {6}^{2} }{9} = \frac{36}{9} = 4$

6 0

4 years ago

Enter an equation for the function. Give your answer in the form a(b)". In the event that a = 1, give your

Anna [14]

Answer:

$r(n)=14,000(0.50)^n$

Step-by-step explanation:

we know that

The equation of a exponential decay function is equal to

$r=a(1-r)^n$

where

r is the number of bacteria remaining

n is the number of hours since the antibiotic was applied

a is the initial value

r is the rate of change

we have

$a=14,000\ bacteria\\r=50\%=50/100=0.50$

substitute

$r(n)=14,000(1-0.50)^n$

$r(n)=14,000(0.50)^n$

8 0

4 years ago

Suppose that you want to estimate the mean pH of rainfalls in an area that suffers from heavy pollution due to the discharge of

labwork [276]

Answer:

The number of rainfalls is $n =96$

The answer to the second question is no it will not be valid this because from the question we are told that the experiment require one pH reading per rainfall so getting multiply specimens(used for the pH reading) from one rainfall will make the experiment invalid.

Step-by-step explanation:

from the question we are told that

The standard deviation is $\sigma = 0.5$

The margin of error is $E = 0.1$

Given that the confidence level is 95% then we can evaluate the level of significance as

$\alpha = 100 - 95$

$\alpha = 5 \%$

$\alpha =0.05$

Next we will obtain the critical value of $\frac{\alpha }{2}$ from the normal distribution table , the value is $Z_{\frac{\alpha }{2} } = 1.96$

Generally the sample size is mathematically represented as

$n = [\frac{Z_{\frac{\alpha }{2} * \sigma }}{ E} ]^2$

substituting values

$n = [\frac{1.96 * 0.5 }{ 0.1} ]^2$

$n =96$

The answer to the second question is no the validity is null this because from the question we are told that the experiment require one pH reading per rainfall so getting multiply specimens(used for the pH reading) from one rainfall will make the experiment invalid

5 0

4 years ago