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telo118 [61]
3 years ago
10

You bake 42 chocolate cupcakes and 28 red velvet cupcakes.You package them in boxes that have the same ratio of chocolate to red

velvet as the total cupcakes.How many red velvet cupcakes are in a box that has 24 chocolate cupcakes?
Mathematics
1 answer:
blondinia [14]3 years ago
5 0

Answer: 16 red velvet cupcakes.

Step-by-step explanation:

To solve this problem you can apply the following proccedure:

- Let's call the number of red velvet cupcakes in a box that has 24 chocolate cupcakes: x

- Based on the information given in the problem, you can set up the following proportion and Solve for x<em>:</em>

\frac{42}{28}=\frac{24}{x}\\\\1.5x=24\\x=16

There are 16 red velvet cupcakes in a box that has 24 chocolate cupcakes.

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Suppose that the mean time that visitors stay at a museum is 94.2 minutes with a standard deviation of 15.5 minutes. The standar
Maurinko [17]

Answer:

94.2 -0.994*3.1 = 91.1186

94.2 +0.994*3.1 = 97.2814

And the 68% confidence interval is given by (91.1186, 97.2814)

Step-by-step explanation:

For this case we know that mean time that visitors stay at a museum is given by:

\bar X = 94.2

The standard deviation is given by:

s= 15.5

And the standard error is given by:

SE = \frac{s}{\sqrt{n}} =3.1

And we want to interval captures 68% of the means for random samples of 25 scores and for this case the critical value can be founded like this using the normal standard distribution or excel:

z_{\alpha/2}= \pm 0.994

We can find the interval like this:

\bar X \pm ME

And replacing we got:

94.2 -0.994*3.1 = 91.1186

94.2 +0.994*3.1 = 97.2814

And the 68% confidence interval is given by (91.1186, 97.2814)

7 0
3 years ago
4-1/x divided by 16-1/xsquared
Vladimir [108]
4 - 1/x (16)-1/x 2  
4x-18/x
3 0
3 years ago
Solve the following equation: lim x-&gt;0 sin3x/5x^3-4x.
goblinko [34]
Usual limit of sin is sinX/X--->1, when X--->0

sin3x/5x^3-4x=0/0?, sin3x/3x--->1 when x --->0, so sin3x/5x^3-4x=                    [3x. sin3x / 3x] /(5x^3-4x)=(sin3x / 3x) . (3x/5x^3-4x)
   =(sin3x / 3x) . (3/5x^2- 4)
finally lim  sin3x/5x^3-4x=lim (sin3x / 3x) .(3/5x^2- 4)=1x(3/-4)= - 3/4
                 x----->0            x---->0
3 0
3 years ago
Find the area of the largest square contained by a circle of radius r = 1cm. Explain your answer and justify that it is correct.
Elena-2011 [213]

Answer:

2 square cm

Step-by-step explanation:

Given :

A square is inscribed in a circle whose radius is r = 1 cm

Therefore, the diameter of the circle is 2 r = 2 x 1

                                                                      = 2 cm.

So the diagonal of the square is 2r.

Using the Pythagoras theorem, we find each of the side of the triangle is $r \sqrt 2$.

Therefore, the area of the square is given by $\text{(side)}^2$

                                                                         = $(r\sqrt 2)^2$

                                                                         $= 2 r^2$

                                                                         $= 2 (1)^2$

                                                                         $=2 \ cm^2$

Hence the area of the largest square that is contained by a circle of radius 1 cm is 2 cm square.

7 0
3 years ago
FindFind the length of the crust(arc) of the pizza
Soloha48 [4]

Answer:

29.3 cm

Step-by-step explanation:

converting 70 degrees into radians, we get=(7/18)×π

now we know,

length of the arc= {(7/18)×π} × 24 (radius of the circle)

= 29.3 cm

8 0
3 years ago
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