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3 years ago

ABC is an isosceles triangle with AB=BC ACD is a straight line work out the size of BCD

Mathematics

1 answer:

rusak2 [61]3 years ago

5 0

Answer:

Triangle ABC is isosceles, with AC = BC. Angle ACD= 620 BCD is a straight line. 180-118=62 62:2-31 x= 31.

Step-by-step explanation:

plz mark as brainiest

hope it helps

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Put into scientific notation 2910000

dolphi86 [110]

Answer:

The Answer is: 2.91 x 10^6

Step-by-step explanation:

Move the decimal from the right to left and count the number of moves, stopping at the next to last number.

Given:

2,910,000

In Scientific notation:

2.91 x 10^6

Hope this helps! Have an Awesome Day!! :-)

7 0

3 years ago

Find the answer to each of the fot

stich3 [128]

Answers:
1 is -8
2 is 29
3 is -15
4 is -1

Hope this answer helps and have a nice day

3 0

3 years ago

Help fast please ! Match the graph go one of the exponential functions

lara [203]

Answer:

$f(x)=(\frac{1}{4} )^{x} +4$

Step-by-step explanation:

The way I did this was I noticed that the graph had a y-intercept of 4 and the only exponential function that had a y-intercept of 4 was $f(x)=(\frac{1}{4} )^{x} +4$

3 0

3 years ago

Let represent the number of tires with low air pressure on a randomly chosen car. The probability distribution of is as follows.

Sindrei [870]

Answer:

a) $P(X=3) = 0.1$

b) $P(X\geq 3) =1-P(X$

And replacing we got:

$P(X \geq 3) = 1- [0.2+0.3+0.1]= 0.4$

c) $P(X=4) = 0.3$

d) $P(X=0) = 0.2$

e) $E(X) =0*0.2 +1*0.3+2*0.1 +3*0.1 +4*0.3= 2$

f) $E(X^2)= \sum_{i=1}^n X^2_i P(X_i)$

And replacing we got:

$E(X^2) =0^2*0.2 +1^2*0.3+2^2*0.1 +3^2*0.1 +4^2*0.3= 6.4$

And the variance would be:

$Var(X0 =E(X^2)- [E(X)]^2 = 6.4 -(2^2)= 2.4$

And the deviation:

$\sigma =\sqrt{2.4} = 1.549$

Step-by-step explanation:

We have the following distribution

x 0 1 2 3 4

P(x) 0.2 0.3 0.1 0.1 0.3

Part a

For this case:

$P(X=3) = 0.1$

Part b

We want this probability:

$P(X\geq 3) =1-P(X$

And replacing we got:

$P(X \geq 3) = 1- [0.2+0.3+0.1]= 0.4$

Part c

For this case we want this probability:

$P(X=4) = 0.3$

Part d

$P(X=0) = 0.2$

Part e

We can find the mean with this formula:

$E(X)= \sum_{i=1}^n X_i P(X_i)$

And replacing we got:

$E(X) =0*0.2 +1*0.3+2*0.1 +3*0.1 +4*0.3= 2$

Part f

We can find the second moment with this formula

$E(X^2)= \sum_{i=1}^n X^2_i P(X_i)$

And replacing we got:

$E(X^2) =0^2*0.2 +1^2*0.3+2^2*0.1 +3^2*0.1 +4^2*0.3= 6.4$

And the variance would be:

$Var(X0 =E(X^2)- [E(X)]^2 = 6.4 -(2^2)= 2.4$

And the deviation:

$\sigma =\sqrt{2.4} = 1.549$

4 0

3 years ago

(04.02 LC)

True [87]

The graph is the horizontal line:

y=3 it has a slope of zero.

Function 2 is

y=3x+1, it is in slope-intercept form, y=mx+b where m is the slope and b is the y=intercept so in this case it has a slope of 3.

So the difference in slopes is 3-0=3

3 0

3 years ago