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3 years ago

Can someone please help me with both of these, I'll appreciate it:)

Mathematics

1 answer:

Kipish [7]3 years ago

7 0

2) the answer is the fourth one because they're both moving together so they'd remain at the same distance

3) also the angles shouldn't have changed. like number two it moves as a whole so nothing changes, so it is choice 3 40 degrees

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I need help plzz someone help meee :)

Eddi Din [679]

Answer:

Table 1 = Proportional

y is 7 times x

Table 2 = Not proportional

Step-by-step explanation:

5 0

3 years ago

Read 2 more answers

Would this be 192? if not can someone please explain how to do it?

Alinara [238K]

Answer:

For anything you have to ask question, right? How can someone understand if you just show the figure

5 0

2 years ago

If f(x) = 2x2 + 2 and g(x) = x2 - 1, find (f - g)(x)I

erastovalidia [21]

9514 1404 393

Answer:

C. x^2 + 3

Step-by-step explanation:

Substitute for f(x) and g(x) and simplify.

(f -g)(x) = f(x) -g(x)

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4 0

2 years ago

Merta reports that 74% of its trains are on time. A check of 60 randomly selected trains shows that 38 of them arrived on time.

kenny6666 [7]

Answer:

No, the on-time rate of 74% is not correct.

Solution:

As per the question:

Sample size, n = 60

The proportion of the population, P' = 74% = 0.74

q' = 1 - 0.74 = 0.26

We need to find the probability that out of 60 trains, 38 or lesser trains arrive on time.

Now,

The proportion of the given sample, p = $\frac{38}{60} = 0.634$

Therefore, the probability is given by:

$P(p\leq 0.634) = [\frac{p - P'}{\sqrt{\frac{P'q'}{n}}}]\leq [\frac{0.634 - 0.74}{\sqrt{\frac{0.74\times 0.26}{60}}}]$

P $(p\leq 0.634) = P[z\leq -1.87188]$

P $(p\leq 0.634) = P[z\leq -1.87] = 0.0298$

Therefore, Probability of the 38 or lesser trains out of 60 trains to be on time is 0.0298 or 2.98 %

Thus the on-time rate of 74% is incorrect.

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3 years ago

It costs $15 for a yearly membership to a movie club at a movie theater. A movie ticket costs $5 for club members and $8 for non

Alona [7]

So what is the question?

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2 years ago