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2 years ago

A Cashier has fifteen $5 bills . How Much does he have in $5 bills

1 answer:

4 0

Lets make a known and an unknown chart:

KNOWN: Fifteen $5 bills.

UNKNOWN: How much is that?

So to find the answer to this, we are going to do 5+5+5+5+5+5+5+5+5+5+5+5+5+5+5=??
The answer is $75.
~Hope I helped!~

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Find the dimensions of the open rectangular box of maximum volume that can be made from a sheet of cardboard 11 in. by 7 in. by

11111nata11111 [884]

Answer:

The dimension of the open rectangular box is $8.216\times 4.216\times 1.392$ .

The volume of the box is 8.217 cubic inches.

Step-by-step explanation:

Given : The open rectangular box of maximum volume that can be made from a sheet of cardboard 11 in. by 7 in. by cutting congruent squares from the corners and folding up the sides.

To find : The dimensions and the volume of the open rectangular box ?

Solution :

Let the height be 'x'.

The length of the box is '11-2x'.

The breadth of the box is '7-2x'.

The volume of the box is $V=l\times b\times h$

$V=(11-2x)\times (7-2x)\times x$

$V=4x^3-36x^2+77x$

Derivate w.r.t x,

$V'(x)=4(3x^2)-2(36x)+77$

$V'(x)=12x^2-72x+77$

The critical point when V'(x)=0

$12x^2-72x+77=0$

Solve by quadratic formula,

$x=\frac{18+\sqrt{93}}{6},\frac{18-\sqrt{93}}{6}$

$x=4.607,1.392$

Derivate again w.r.t x,

$V''(x)=24x-72$

Now, $V''(4.607)=24(4.607)-72=38.568>0$ (+ve)

$V''(1.392)=24(1.392)-72=-38.592$ (-ve)

So, there is maximum at x=1.392.

The length of the box is $l=11-2x$

$l=11-2(1.392)=8.216$

The breadth of the box is $b=7-2x$

$b=7-2(1.392)=4.216$

The height of the box is h=1.392.

The dimension of the open rectangular box is $8.216\times 4.216\times 1.392$ .

The volume of the box is $V=l\times b\times h$

$V=8.216\times 4.216\times 1.392$

$V=48.217\ in.^3$

The volume of the box is 8.217 cubic inches.

5 0

3 years ago

In a school of 330 students, 85 of them are in the drama club, 200 of them are in a sports team and 60 students do drama and spo

PilotLPTM [1.2K]

Answer:

nn n h hbjfbyfdxjnw1urdnbr3rb3j ruh3w drk4jtYHU&jt5krde,frtnjhd,dbf

Step-by-step explanation:

jddddddmk.aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa

7 0

3 years ago

I have a past due balance of $87.50 and my new charges are $75.00. I was also charged a late fee of 15% of my past due balance.

Kitty [74]

Past due balance = $87.50
Late fee = 15% of $87.50 = 0.15*87.50 = $13.125
New charges = $75.00

New total = 87.50 + 13.125 + 75 = $175.625

Answer: The new total is $175.63

8 0

3 years ago

Read 2 more answers

Solve each equation.<br> Show all steps<br> 4) -8(-6-5k)=-232

Lubov Fominskaja [6]

the answer is k=4.6. hope this helps

6 0

3 years ago

The mean SAT score in mathematics, M, is 600. The standard deviation of these scores is 48. A special preparation course claims

kupik [55]

Answer:

Step-by-step explanation:

The mean SAT score is $\mu=600$ , we are going to call it \mu since it's the "true" mean

The standard deviation (we are going to call it $\sigma$ ) is

$\sigma=48$

Next they draw a random sample of n=70 students, and they got a mean score (denoted by $\bar x$ ) of $\bar x=613$

The test then boils down to the question if the score of 613 obtained by the students in the sample is statistically bigger that the "true" mean of 600.

- So the Null Hypothesis $H_0:\bar x \geq \mu$

- The alternative would be then the opposite $H_0:\bar x < \mu$

The test statistic for this type of test takes the form

$t=\frac{| \mu -\bar x |} {\sigma/\sqrt{n}}$

and this test statistic follows a normal distribution. This last part is quite important because it will tell us where to look for the critical value. The problem ask for a 0.05 significance level. Looking at the normal distribution table, the critical value that leaves .05% in the upper tail is 1.645.

With this we can then replace the values in the test statistic and compare it to the critical value of 1.645.

$t=\frac{| \mu -\bar x |} {\sigma/\sqrt{n}}\\\\= \frac{| 600-613 |}{48/\sqrt(70}}\\\\= \frac{| 13 |}{48/8.367}\\\\= \frac{| 13 |}{5.737}\\\\=2.266\\$

<h3>since 2.266>1.645 we can reject the null hypothesis.</h3>

6 0

3 years ago

Read 2 more answers