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riadik2000 [5.3K]
2 years ago
10

HElp And THAnkS ASAP

Mathematics
1 answer:
Kobotan [32]2 years ago
6 0

Answer: i’m not sure on which one it might be but i’m guessing a or c

Step-by-step explanation:

i’m guessing a or c cause we tryna figure it out only for jeanie

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Need help with questions 9,10 and 11 from the bottom<br><br> Will give brainliest!
qwelly [4]

Step-by-step explanation:

9:

A=V/t

A=8.0m/s•1/2.0s

A=8.0m/2.0s²

A=4.0m/s²

10:

V=A•time

V=4.0m/s²•2.3s

V=9.2m/s

11:

V=A•t

24.0=5.6t

4.3s=t

5 0
3 years ago
Read 2 more answers
Show 2 different solutions to the task.
laila [671]

Answer with Step-by-step explanation:

1. We are given that an expression n^2+n

We have to prove that this expression is always is even for every integer.

There are two cases

1.n is odd integer

2.n is even integer

1.n is an odd positive integer

n square is also odd integer and n is odd .The sum of two odd integers is always even.

When is negative odd integer then n square is positive odd integer and n is negative odd integer.We know that difference of two odd integers is always even integer.Therefore, given expression is always even .

2.When n is even positive integer

Then n square is always positive even integer and n is positive integer .The sum of two even integers is always even.Hence, given expression is always even when n is even positive integer.

When n is negative even integer

n square is always positive even integer and n is even negative integer .The difference of two even integers is always even integer.

Hence, the given expression is always even for every integer.

2.By mathematical induction

Suppose n=1 then n= substituting in the given expression

1+1=2 =Even integer

Hence, it is true for n=1

Suppose it is true for n=k

then k^2+k is even integer

We shall prove that it is true for n=k+1

(k+1)^1+k+1

=k^1+2k+1+k+1

=k^2+k+2k+2

=Even +2(k+1)[/tex] because k^2+k is even

=Sum is even because sum even numbers is also even

Hence, the given expression is always even for every integer n.

3 0
3 years ago
From a large number of actuarial exam scores, a random sample of scores is selected, and it is found that of these are passing s
Mnenie [13.5K]

<u>Supposing 60 out of 100 scores are passing scores</u>, the 95% confidence interval for the proportion of all scores that are passing is (0.5, 0.7).

  • The lower limit is 0.5.
  • The upper limit is 0.7.

In a sample with a number n of people surveyed with a probability of a success of \pi, and a confidence level of \alpha, we have the following confidence interval of proportions.

\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}

In which

z is the z-score that has a p-value of \frac{1+\alpha}{2}.

60 out of 100 scores are passing scores, hence n = 100, \pi = \frac{60}{100} = 0.6

95% confidence level

So \alpha = 0.95, z is the value of Z that has a p-value of \frac{1+0.95}{2} = 0.975, so z = 1.96.

The lower limit of this interval is:

\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.6 - 1.96\sqrt{\frac{0.6(0.4)}{100}} = 0.5

The upper limit of this interval is:

\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.6 + 1.96\sqrt{\frac{0.6(0.4)}{100}} = 0.7

The 95% confidence interval for the proportion of all scores that are passing is (0.5, 0.7).

  • The lower limit is 0.5.
  • The upper limit is 0.7.

A similar problem is given at brainly.com/question/16807970

5 0
2 years ago
HELPPPPPPPPPPPPPPPPPPPPPPP solving quadratics with the Calculator.
Dima020 [189]
Well I guess and I hope it helped
3 0
3 years ago
Solve: the sum of two numbers is 70. one number is 4 times the other. what are the two numbers?
Thepotemich [5.8K]
Lets say two numbers are x and y

x+y=70-----------------------eq1

x=4y--------------------eq2

4y+y=70
5y=70
y=70/5
y=14 

then x=4*14

x=56


8 0
3 years ago
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