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katovenus [111]
3 years ago
6

Find the product of 51 and -2.5

Mathematics
2 answers:
bearhunter [10]3 years ago
5 0

Answer:

-127.5

you have to multiply -2.5 with 1

then -2.5 with 50

then add

you get -127.5

Elodia [21]3 years ago
3 0

Answer: -127.5

Hope this helps!

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The accompanying data represent the daily​ (for example, Monday to​ Tuesday) movement of Johnson​ & Johnson​ (JNJ) stock for
egoroff_w [7]

Supposing that the stock increases in 37 days, the 95% confidence interval for the proportion of days JMJ stock increases is: (0.484, 0.7292)

  • The lower bound is of 0.484.
  • The upper bound is of 0.7292.
  • The interpretation is that we are <u>95% sure that the true proportion</u> of all days in which the JMJ stock increases <u>is between 0.484 and 0.7292.</u>

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}.

Supposing that it increases on 37 out of 61 days:

n = 61, \pi = \frac{37}{61} = 0.6066

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}} = 0.6066 - 1.96\sqrt{\frac{0.6066(0.3934)}{61}} = 0.484

The upper limit of this interval is:

\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.6066 + 1.96\sqrt{\frac{0.6066(0.3934)}{61}} = 0.7292

The ​95% confidence interval for the proportion of days JMJ stock increases is (0.484, 0.7292), in which 0.484 is the lower bound and 0.7292 is the upper bound.

The interpretation is that we are <u>95% sure that the true proportion</u> of all days in which the JMJ stock increases <u>is between 0.484 and 0.7292.</u>

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

4 0
2 years ago
BRAINLIESTTTT ASAP! PLEASE ANSWER
iren [92.7K]
H(t) = −16t^2 + 75t + 25
g(t) = 5 + 5.2t

A)
At 2, h(t) = 111, g(t) = 15.4
At 3, h(t) = 106, g(t) = 20.6
At 4, h(t) = 69, g(t) = 25.8
At 5, h(t) = 0, g(t) = 31
The heights of both functions would have been the closest value to each other after 4 seconds, but before 5 seconds. This is when g(x) is near 30 (26-31), and the only interval that h(t) could be near 30 is between 4 and 5 seconds (as it is decreasing from 69-0).

B) The solution to the two functions is between 4 and 5 seconds, as that is when their height is the same for both g(t) and h(t). Actually the height is at 4.63 seconds, their heights are both
What this actually means is that this time and height is when the balls could collide; or they would have hit each other, given the same 3-dimensional (z-axis) coordinate in reality.

6 0
3 years ago
Read 2 more answers
a student showed the steps below while solving the equation 14=log5(2x-3) by graphing. which step did the student make the 1sr e
mixer [17]

Answer:

x= \frac{5^{14}+3}{2}

Step-by-step explanation:

The correct steps to solve the equation are:

14=log_5(2x-5)

5^{14}=5^{log_5(2x-3)}

Because a^{log_am}=m

So, solving we get:

5^{14}=2x-3

Sum 3 on every side:

5^{14}+3=2x-3+3\\5^{14}+3=2x

Dividing by 2 into both sides:

\frac{5^{14}+3}{2}=\frac{2x}{2}\\\frac{5^{14}+3}{2}=x

So, the answer is  x= \frac{5^{14}+3}{2}

7 0
3 years ago
I’m completely lost. Answer?
aliya0001 [1]
I think it's true because its the same data set just put in order backwards.


Hope this helps!! :-)
3 0
3 years ago
Find the 75th term of the following6, 10, 14, 18,
enot [183]

Given the sequence:

6, 10, 14, 18,...

We will find the 75th term

The given sequence is an arithmetic sequence

Because there is a constant common differnce

d = 18 - 14 = 14 - 10 = 10 - 6 = 4

The first term = a = 6

The general formula of the arithmetic sequence is as follows:

a_n=a+d(n-1)

Where: n is the nth term

To find the 75th term, substitute with n = 75 and a = 6, d = 4

a_{75}=6+4\cdot(75-1)=302

So, the answer will be the 75th term = 302

7 0
1 year ago
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