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schepotkina [342]

3 years ago

6

You spin the spinner twice what is the probability of landing on a three and then landing on a number greater than three

1 answer:

timama [110]3 years ago

3 0

Answer:

2/4

Step-by-step explanation:

You might be interested in

What is the answer to this word problem??

AveGali [126]

Let’s use m for months
So,
1st service
10m + 20

1. 10 + 20 =30
2. 10x2=20+20=40
3.10x3=30+20=50
4.10x4=40+20=60
5.10x5=50+20=70

2nd service
11m + 15

1. 11+15=26
2. 11x2=22+15=37
3 11x3=33+15=48
4.11x4=44+15=59
5.11x5=55+15=70

So, the 5th month the services cost the same

Hope this helps! ;)

7 0

3 years ago

Choose the simplified form of the fifth term of the expansion. -60x2y8 -30x2y8 30x2y8 60x2y8

Vika [28.1K]

Answer:

Answer is D on edge

Step-by-step explanation:

5 0

3 years ago

Read 2 more answers

X + 7<br> 2x - 3<br> Perimeter:<br> Area:

777dan777 [17]

Answer:

P=6x+8

A=2x^2+11x-21

Step-by-step explanation:

Perimeter:

P=2(x+7)+2(2x-3)

P=6x+8

Area:

A=(x+7)(2x-3)

A=2x^2+11x-21

7 0

4 years ago

8_murik_8 [283]

Answer:

alternate exterior and transversal P

consecutive interior angles and transversal R

corresponding and transversal S

alternate interior and transversal Q

5 0

3 years ago

A product is made up of components A, B, C, D, E, F, G, H, I, and J. Components A, B, C, and F have a 1/10,000 chance of failure

lesantik [10]

Answer:

The overall reliability is 99.7402 %

Step-by-step explanation:

The overall reliability of the product is calculated as the product of the working probability of the components.

For components A,B,C and F we have :

$P(failure)=\frac{1}{10000}$

⇒

$P(Work)=1-\frac{1}{10000}=\frac{9999}{10000}=0.9999$

For components D,E,G and H we have :

$P(failure)=\frac{3}{10000}$

⇒

$P(Work)=1-\frac{3}{10000}=\frac{9997}{10000}=0.9997$

Finally, for components I and J :

$P(failure)=\frac{5}{10000}$

⇒

$P(Work)=1-\frac{5}{10000}=\frac{9995}{10000}=0.9995$

Now we multiply all the working probabilities. We mustn't forget that we have got ten components in this case :

Components A,B,C and F with a working probability of 0.9999

Components D,E,G and H with a working probability of 0.9997

Components I and J with a working probability of 0.9995

Overall reliability = $(0.9999)^{4}(0.9997)^{4}(0.9995)^{2}=0.997402$

0.997402 = 99.7402 %

4 0

3 years ago