An algebra 2 question need help with....

A company buys pens at the rate of $5.50 per box for the first 10 boxes, $4.50 per box for the next 10 boxes, and $3.50 per box

marusya05 [52]

So it is really easy to solve firstly we can see how much does the first 10 boxes make which makes around 55$ obviously. Secondly 45$ for the next 10 boxes.

So for now we can simply calculate that we have spent around 100$ which means 20 boxes. The remaining money left is 77$ so we can buy 77/3.5 = 22 only 22 boxes with that money. Hence a total of 42 boxes.

4 0

4 years ago

Please help.. thanks if you do :)

AlladinOne [14]

Answer:

it X+3

it 3 hope this help yuuu

Step-by-step explanation: 6(x+2)=30

6x+6.2=5

6(x+2)=30

6x + 12 = 30

-12 -12

6x = 18

x = 3

hope this help the best answer is X=3

tht the bnest answer for the q

6 0

3 years ago

Read 2 more answers

Scores on the SAT Mathematics test are believed to be normally distributed. The scores of a simple random sample of five student

AysviL [449]

Answer:

The mean calculated for this case is $\bar X=584$

And the 95% confidence interval is given by:

$584-2.776\frac{86.776}{\sqrt{5}}=476.271$

$584+2.776\frac{86.776}{\sqrt{5}}=691.729$

So on this case the 95% confidence interval would be given by (476.271;691.729)

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

$\bar X$ represent the sample mean for the sample

$\mu$ population mean (variable of interest)

s represent the sample standard deviation

n represent the sample size

Solution to the problem

The confidence interval for the mean is given by the following formula:

$\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}$ (1)

In order to calculate the mean and the sample deviation we can use the following formulas:

$\bar X= \sum_{i=1}^n \frac{x_i}{n}$ (2)

$s=\sqrt{\frac{\sum_{i=1}^n (x_i-\bar X)}{n-1}}$ (3)

The mean calculated for this case is $\bar X=584$

The sample deviation calculated $s=86.776$

In order to calculate the critical value $t_{\alpha/2}$ we need to find first the degrees of freedom, given by:

$df=n-1=5-1=4$

Since the Confidence is 0.95 or 95%, the value of $\alpha=0.05$ and $\alpha/2 =0.025$ , and we can use excel, a calculator or a tabel to find the critical value. The excel command would be: "=-T.INV(0.025,4)".And we see that $t_{\alpha/2}=2.776$