Please help me! will mark brainliest

What is wrong with this problem. 2(n +3) = 2n +3

LekaFEV [45]

Steps to solve:

2(n + 3) = 2n + 3

~Distribute left side

2n + 6 = 2n + 3

~Subtract 6 to both sides

2n = 2n - 3

~Subtract 2n to both sides

n = -3

Best of Luck!

7 0

3 years ago

Name; mia

Reika [66]

Answer:

you need to add and then subtract and the anw=sewr wi;l be 14567890 .happy and boom your done

Step-by-step explanation:

3 0

3 years ago

50 points! I understand A. and B. but I would really appreciate help with C.

FromTheMoon [43]

Answer:

$51.72\text{ cells per hour}$

Step-by-step explanation:

So, the function, P(t), represents the number of cells after t hours.

This means that the derivative, P'(t), represents the instantaneous rate of change (in cells per hour) at a certain point t.

C)

So, we are given that the quadratic curve of the trend is the function:

$P(t)=6.10t^2-9.28t+16.43$

To find the <em>instanteous</em> rate of growth at t=5 hours, we must first differentiate the function. So, differentiate with respect to t:

$\frac{d}{dt}[P(t)]=\frac{d}{dt}[6.10t^2-9.28t+16.43]$

Expand:

$P'(t)=\frac{d}{dt}[6.10t^2]+\frac{d}{dt}[-9.28t]+\frac{d}{dt}[16.43]$

Move the constant to the front using the constant multiple rule. The derivative of a constant is 0. So:

$P'(t)=6.10\frac{d}{dt}[t^2]-9.28\frac{d}{dt}[t]$

Differentiate. Use the power rule:

$P'(t)=6.10(2t)-9.28(1)$

Simplify:

$P'(t)=12.20t-9.28$

So, to find the instantaneous rate of growth at t=5, substitute 5 into our differentiated function:

$P'(5)=12.20(5)-9.28$

Multiply:

$P'(5)=61-9.28$

Subtract:

$P'(5)=51.72$

This tells us that at <em>exactly</em> t=5, the rate of growth is 51.72 cells per hour.

And we're done!

7 0

3 years ago

A random sample of 16 one-kilogram sugar packets is obtained and the actual weights of the packets are measured. The sample mean

elena55 [62]

Answer:

The 99% two-sided confidence interval for the average sugar packet weight is between 0.882 kg and 1.224 kg.

Step-by-step explanation:

We are in posession of the sample's standard deviation, so we use the student's t-distribution to find the confidence interval.

The first step to solve this problem is finding how many degrees of freedom, we have. This is the sample size subtracted by 1. So

df = 16 - 1 = 15

99% confidence interval

Now, we have to find a value of T, which is found looking at the t table, with 35 degrees of freedom(y-axis) and a confidence level of $1 - \frac{1 - 0.99}{2} = 0.905([tex]t_{995}$ ). So we have T = 2.9467

The margin of error is:

M = T*s = 2.9467*0.058 = 0.171

In which s is the standard deviation of the sample.

The lower end of the interval is the sample mean subtracted by M. So it is 1.053 - 0.171 = 0.882kg

The upper end of the interval is the sample mean added to M. So it is 1.053 + 0.171 = 1.224 kg.

The 99% two-sided confidence interval for the average sugar packet weight is between 0.882 kg and 1.224 kg.

3 0

3 years ago

Adam bought A new pair of baseball cleats for $65 the sales tax is 5.5% what is the total cost in dollars that Adam will pay

Dmitry [639]

(65×5.5%)+65 = 68.575

8 0

4 years ago