Find the GCF of 56 and 46.

What is the area of the real object that the scale drawing models?

Kruka [31]

Answer:

HOPE THIS HELPS!!

Step-by-step explanation:

YOU could find it by doing these steps:

Atual Area = �� square units.

Scale Drawing Area = �� square units.

Value of the Ratio of the Scale Drawing Area to the Actual Area:

7 0

3 years ago

Tia is making lemonade for a baseball tournament. The recipe calls for 2/3 cup of strawberries for each gallon of lemonade. She

antoniya [11.8K]

Given:

The recipe calls for $\dfrac{2}{3}$ cup of strawberries for each gallon of lemonade.

Tia has 6 cups of strawberries.

To find:

The number of gallons of lemonade she can make.

Solution:

We have,

$\dfrac{2}{3}$ cup of strawberries for = 1 gallon of lemonade

$1$ cup of strawberries for = $\dfrac{1}{\frac{2}{3}}$ gallons of lemonade

= $1\times \dfrac{3}{2}$ gallons of lemonade

= $\dfrac{3}{2}$ gallons of lemonade

$6$ cup of strawberries for = $6\times \dfrac{3}{2}$ gallons of lemonade

= $\dfrac{18}{2}$ gallons of lemonade

= $9$ gallons of lemonade

Therefore, 9 gallons of lemonade she can make by using 6 cups of strawberries.

7 0

2 years ago

A store randomly samples 603 shoppers over the course of a year and finds that 142 of them made their visit because of a coupon

fiasKO [112]

Answer:

The 95% confidence interval for the fraction of all shoppers during the year whose visit was because of a coupon they'd received in the mail is (0.2016, 0.2694).

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of $\pi$ , and a confidence level of $1-\alpha$ , we have the following confidence interval of proportions.

$\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}$

In which

z is the z-score that has a p-value of $1 - \frac{\alpha}{2}$ .

A store randomly samples 603 shoppers over the course of a year and finds that 142 of them made their visit because of a coupon they'd received in the mail.

This means that $n = 603, \pi = \frac{142}{603} = 0.2355$

95% confidence level

So $\alpha = 0.05$ , z is the value of Z that has a p-value of $1 - \frac{0.05}{2} = 0.975$ , so $Z = 1.96$ .

The lower limit of this interval is:

$\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.2355 - 1.96\sqrt{\frac{0.2355*0.7645}{603}} = 0.2016$

The upper limit of this interval is:

$\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.2355 + 1.96\sqrt{\frac{0.2355*0.7645}{603}} = 0.2694$

The 95% confidence interval for the fraction of all shoppers during the year whose visit was because of a coupon they'd received in the mail is (0.2016, 0.2694).

8 0

2 years ago

More on Areas. Farmer Jones, and his wife, Dr. Jones, both mathematicians, decide to build a fence in their field to keep the sh

Olin [163]

Answer:

Area covered by the fences will be 16.1 unit²

Step-by-step explanation:

Let the first parabola is represented by the function f(x) = 6x²

and second parabola by g(x) = x² + 9

point of intersection of the graphs will be determined when f(x) = g(x)

6x² = x² + 9

5x² = 9

x² = 1.8

x = ± 1.34

Now we will find the area between these curves drawn on the graph.

Area = $\int_{-1.34}^{1.34}[f(x)-g(x)]dx=\int_{-1.34}^{1.34}[6x^{2}-(x^{2}+9)]dx$

= $\int_{-1.34}^{1.34}(5x^{2}-9)dx$

= $[\frac{5}{3}x^{3}-9x]_{-1.34}^{1.34}$

= $[\frac{5}{3}(-1.34)^{3}-9(-1.34)-\frac{5}{3}(1.34)^{3}+9(1.34)]$

= $[-4.01+12.06-4.01+12.06]$

= 16.1 unit²

6 0

3 years ago

Find the product of z1 and z2, where z1 = 7(cos 40° + i sin 40°) and z2 = 6(cos 145° + i sin 145°).

Oduvanchick [21]

For two complex numbers $z_1=re^{i\theta}=r(\cos\theta+i\sin\theta)$ and $z_2=se^{i\varphi}=s(\cos\varphi+i\sin\varphi)$ , the product is

$z_1z_2=rse^{i(\theta+\varphi)}=rs(\cos(\theta+\varphi)+i\sin(\theta+\varphi))$

That is, you multiply the moduli and add the arguments. You have $z_1=7e^{i40^\circ}$ and $z_2=6e^{i145^\circ}$ , so the product is

$z_1z_2=7\times6(\cos(40^\circ+145^\circ)+i\sin(40^\circ+145^\circ)=42(\cos185^\circ+i\sin185^\circ)=42e^{i185^\circ}$

3 0

3 years ago

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