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Lady bird [3.3K]
3 years ago
6

Help Please(pre algebra)

Mathematics
1 answer:
miv72 [106K]3 years ago
7 0
Where’s the problem?
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The sequence is <br> 4 1 -2 -5 -8<br> What is the nth term
UkoKoshka [18]

Answer:

subtract 3 each term

Step-by-step explanation:

8 0
3 years ago
Find x and y, given that line WS and line VT are parallel. Show all work!!
slavikrds [6]
Given that those lines are parallel they will form similar traingles. This means the lengths of the sides of the triangles will be proportional through a ratio.

Find the ratio:  divde the given small side by the given big side(they must be the same sides of the triangles though)

For example:
I found the ratio by dividing 8 / (20) and simplify to 2/5

Next step: multiply 22 by the fraction to get "y"
\frac{y}{22} = \frac{2}{5}
22*\frac{2}{5}= y
y=8.8 units


To find x :

Step 1: write an equation for the length of side US.
US= 5+x-2= x+3

Step 2: DIVIDE 5 by the ratio (must divide to get longer side DO NOT multiply)
US= 5/(2/5)= 12.5 units

Step 3: set the expression you found for US equal to 12.5 

12.5= x+3
subtract 3 from both sides to isolate x.
x= 9.5 units

My answers:
x= 9.5 units
y= 8.8 units
5 0
3 years ago
Rick and Ranger sell Toyotas for Tom's Auto. Over the past year they sold 360 cars. Assuming Rick sells 5 times as many as Range
gulaghasi [49]
360/6 = 60

Rick = 60 x 5 = 300 so Ranger = 360 - 300 = 60

Ranger sold 60 cars
6 0
3 years ago
There are 20 girls and 15 boys in the class. How many different five-member teams could be formed if each team should be compose
Paul [167]

The number of combinations which are possible according to the specifications is; 119,700.

<h3>How many combinations are there to compose 3 girls and 2 boys?</h3>

It follows from the task content that the 5ember teams to be formed must contain 3 girls and 2 boys.

On this note, it follows that the combinations possible are as follows

20C(3) × 15C2 = 1140 × 105 = 119,700.

Read more on combinations;

brainly.com/question/11732255

#SPJ1

4 0
2 years ago
Late shows Some TV shows begin after their scheduled times when earlier programs run late. According to a network’s records, abo
gladu [14]

Answer:

No because the probability of consecutive shows starting late are not independent events

Step-by-step explanation:

Is good begin with the definition of independent events

When we say Independent Events we are refering to  events which occur with no dependency of other evnts. Basically when the occurrence of one event is not affected by another one.

When two events are independent P(A and B) = P(A)xP(B)

But for this case we can't multiply 0.97x0.97x0.97 in order to find the probability that 3 consecutive shows start on time, because the probability for shows starting late are not independent events, because if the second show is late, the probability that the next show would be late is higher. And for this reason we can use the independency concept here and the multiplication of probabilities in order to find the probability required.

3 0
3 years ago
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