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2 years ago

Pls answer the question and show how or why

Mathematics

2 answers:

postnew [5]2 years ago

4 0

Answer:

there is no question my dude

Step-by-step explanation:

andrew-mc [135]2 years ago

3 0

What question sweetie?

You might be interested in

We are interested in the dimensions of a certain rectangle. this rectangle has length twice the side of the square and width thr

Mila [183]

Area of square = s^2
Area of Rectangle = lw
l = 2s

w+3 = s
solve for w, w=s-3
Now we have l and w.
Plug both into area of rectangle formula so: (2s)(s-3)
Since both areas are equal set both equations equal to each other:
(2s)(s-3)=s^2
Now simplify
2s^2-6s=s^2
s^2-6s=0, Solve for s.
Factor polynomial. s(s-6)=0 , s can be equal to 0 or 6. HOWEVER, you cannot have a side length of 0 therefore the side length has to be 6.
Now plug in s for the length formula for the rectangle:
l = 2s so... l = 2(6) so length of rectangle = 12.
Now plug in s for the width formula for the rectangle:
w+3=s so... w+3=6 so width of rectangle = 3.
Now the dimensions of the rectangle are 12 by 3. 12 being length and 3 width.
To CHECK:
Find area of rectangle:
A=lw so A=3 times 12 so A=36
Find area of square:
We know the side is equal to 6 so
A=s^2 so 6^2 = 36
The areas are equal that verifies the answer of 12 by 3.

8 0

3 years ago

This famous clock tower has a clock face with a diameter of 23 feet. What is the approximate area of this clock face

pashok25 [27]

Answer:

Therefore the approximate area of this clock face is 415.3 feet².

Step-by-step explanation:

Given:

Clock is of Circular Shape,

Diameter = 23 feet

$Radius =r =\dfrac{diameter}{2}=\dfrac{23}{2}=11.5\ feet$

To Find :

Area of Clock Face = ?

Solution:

Area of Circle having Radius 'r' is given as,

$\textrm{Area of Circle}=\pi r^{2}$

Substituting the values we get

$\textrm{Area of Clock Face}=3.14\times (11.5)^{2}=415.265=415.3\ feet^{2}$

Therefore the approximate area of this clock face is 415.3 feet².

7 0

3 years ago

What is the Volume if the Mass = 35g and Density = 5g/cm^3 ?

VLD [36.1K]

density = mass / volume

4 = 35 / v

v = 35/4 = 8.75 cm^3

6 0

2 years ago

1) If the alpha level is changed from α = .05 to α = .01, what happens to boundaries for the critical region?

Alexxandr [17]

Answer:

1) a. Move farther into the tails

2) a. Decreases

Step-by-step explanation:

Hello!

Let's say for example that you are making a confidence interval for the mean, using the Z-distribution:

X[bar] ± $Z_{1-\alpha /2}$ * $\frac{Sigma}{\sqrt{n} }$

Leaving all other terms constant, this are the Z-values for three different confidence levels:

90% $Z_{0.95}= 1.648$

95% $Z_{0.975}= 1.965$

99% $Z_{0.995}= 2.586$

Semiamplitude of the interval is

d= $Z_{1-\alpha /2}$ * $\frac{Sigma}{\sqrt{n} }$

Then if you increase the confidence level, the value of Z increases and so does the semiamplitude and amplitude of the interval:

↑d= ↑ $Z_{1-\alpha /2}$ * $\frac{Sigma}{\sqrt{n} }$

They have a direct relationship.

So if you change α: 0.05 to α: 0.01, then the confidence level 1-α increases from 0.95 to 0.99, and the boundaries move farther into the tails.

The significance level of a hypothesis test is the probability of committing a Type I error.

If you decrease the level from 5% to 1%, then logically, the probability decreases.

I hope this helps!

5 0

3 years ago

Which equation does the graph represent (see picture)

egoroff_w [7]

Y=Mx +b

B is 0 because that’s where the graph intercepts the y axis.

M is found by picking any two points and finding the rise and run because m is equal rise/run

Points of choice: (-2,4) and (2,-4)

=2−1/2−1 = -8/4

m is equal -2

Y= -2X

Look at the picture below to differentiate between positive and negative slopes.

8 0

3 years ago