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3 years ago

Regular quadrilateral pyramid with base edge b=55 mm and lateral edge l=7 cm. Find the sum of all edges

2 answers:

4 0

The correct answer is 50 cm or 500 mm. There are 4 base edges and 4 lateral edges, so just multiply the number for both edges by 4 and add. The question threw a curveball at us by giving us two units of measurement, millimeters and centimeters, but since we know that there are 10 millimeters in a centimeter, we can just divide 55 by 10 to get 5.5.

5.5 x 4 = 22
7 x 4 = 28
28 + 22 = 50

If it helps, you can draw the figure and label the edges. Remember that the lateral edge of a quadrilateral pyramid is the slanted line leading up to the top of the pyramid, and that the base edges are the sides of the quadrilateral on the bottom.

Hope this helps!

AlladinOne [14]3 years ago

3 0

Answer:

Step-by-step explanation:

55 · 4 - 10 · 7 = 220 - 70

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2 years ago

The graph below shows the number of paintballs a machine launches, y, in x seconds:

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The expression that can be used to calculate the rate per second at which the machine launches the balls is fraction 20 over 4.

Step-by-step explanation:

The x-axis label is Time in seconds, and the x-axis values are from 0 to 20 in increments of 4 for each grid line.

And the y-axis label is Number of Balls and the y-axis values from 0 to 100 in increments of 20 for each grid line.

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Therefore, the expression that can be used to calculate the rate per second at which the machine launches the balls is fraction 20 over 4. (Answer)

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3 years ago

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Which value of a in the exponential function below would cause the function to shrink? f(x) = a(three-halves) Superscript x Four

Natalka [10]

Answer:

Four-fifths

Step-by-step explanation:

As we know that

By multiplying the function by a constant, we may expand or shrink the function in the y-direction.

Now we have

$y=a(b^{x})$

if $a> 1$ > the function would enlarge

if $0< a < 1$ > the function would shrinks

Now

For case A

$a = \frac{4}{5}$

$0 < (\frac{4}{5} )< 1$ ....... > the function would shrinks

For case B

$a = \frac{5}{4}$

$(\frac{5}{4} )>1$ .......> the function would enlarge

For case C

$a = \frac{3}{2}$

$(\frac{3}{2} )>1$ .........> the function would enlarge

For case D

$a = \frac{7}{4}$

$(\frac{7}{4} )>1$ .........> the function would enlarge

Therefore the second option is correct

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2 years ago

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Answer:

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Step-by-step explanation:

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2 years ago

Three balanced coins are tossed independently. One of the variables of interest is Y1, the number of heads. Let Y2 denote the am

sesenic [268]

Answer:

a) the joint probability distribution function is

P(y1,y2)=

C(3-y1,y2-1)/8, for 1≤y2≤4-y1 and 1≤y1≤3 ,

1/8 for (y1,y2)=(-1 ,0)

0 otherwise

b) the probability is 50% (F(2,1)= 1/2)

Step-by-step explanation:

for the variable Y1, we start with the negative binomial distribution, with true or false values:

p(r,x)=C(x+r-1,r-1)*p^r*(1-p)^x , where x= number of false experiments until a number r of true values is achieved, and C(a,b)=number of combinations of b in a values. if we set r=1 (fist head) and y1=x+1 (except of y1=-1)

p(y1)= C(y1,0)*p^0*(1-p)^(y1-1) = p*(1-p)^(y1-1)

for the variable Y2, we begin with the binomial distribution:

p(n,x)= C(n,x)*p^x*(1-p)^(n-x) , n= number of times the coin is tossed, x= number of heads obtained , p = probability of obtaining heads when the coin is tossed once.

if we set n=3-y1 (remaining coins that can be tail or head), and x=y2-1 (since we already know the first head appeared), we calculate the probability of p(y2) in case of y1 ( or p(y2|y1) )

p(y2|y1)= C(3-y1, y2-1)*p^(y2-1)*(1-p)^(4-y1-y2)

Now using the theorem of Bayes:

P(y2,y1)=p(y2∩y1)=p(y2|y1)*p(y1) = C(3-y1,y2-1)*p^y2*(1-p)^(3-y2)

since the coins are balanced p= 0,5

P(y2,y1)= C(3-y1,y2-1)*0,5^y2*(1-0,5)^(3-y2)= C(3-y1,y2-1)*0,5^3=C(3-y1,y2-1)/8

therefore

P(y2,y1)=

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1/8 for (y1,y2)=(-1 ,0)

0 otherwise

b) F(2,1)= P(y=1,y2<2)+P(y=(-1),y2=0)= [C(2,0)/8+C(2,1)/8] + 1/8 = (1/8+2/8)+1/8=4/8=1/2

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2 years ago