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alexgriva [62]
2 years ago
7

A runner ran 1 4/5 miles on Monday and 6 3/10 on Tuesday. How many times her Monday's distance was her Tuesday's distance? Pleas

e help me I need an answer that makes sense
Mathematics
1 answer:
Tamiku [17]2 years ago
8 0
He ran more than enough that’s all extra talk
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The function f(t)=-5t^2+15t+50 models the approximate height in feet of an object t seconds after it is launched. how many secon
Cloud [144]
F(t) can be factored to find the zeros.
.. f(t) = -5(t -5)(t +2) . . . . f(t) = 0 for t=-2, t=5

The object hits the ground after 5 seconds.
4 0
3 years ago
Help me please what do they mean?
gogolik [260]
Ask google it will help alot
6 0
3 years ago
A random sample of 747 obituaries published recently in Salt Lake City newspapers revealed that 344 (or 46%) of the decedents di
cupoosta [38]

Answer:

The probability value is almost equal to 0. Implying that the proportion of people dying  in that particular interval if deaths occurred randomly throughout the year is unusual.

Step-by-step explanation:

The random variable <em>X</em> can be defined as the number of decedents who died in the three-month period following their birthdays.

A random sample of 747 obituaries published recently in Salt Lake City newspapers revealed that 344 (or 46%) of the decedents died in the three-month period following their birthdays (123).

The probability (p) of anyone dying in any quarter if people die randomly during the year is simply 0.25.

The random variable <em>X</em> follows a Binomial distribution with parameters n = 747 and p = 0.25.

But the sample selected is too large and the probability of success is small.

So a Normal approximation to binomial can be applied to approximate the distribution of <em>p</em> if the following conditions are satisfied:

  1. np ≥ 10
  2. n(1 - p) ≥ 10

Check the conditions as follows:

 np=747\times 0.25=186.75>10\\n(1-p)=747\times (1-0.46)=560.25>10

Thus, a Normal approximation to binomial can be applied.

So,  p\sim N(\hat p,\ \frac{\hat p(1-\hat p)}{n}).

Compute the probability that 46% or more would die in that particular interval if deaths occurred randomly throughout the year as follows:

P (p\geq 0.46)=P(\frac{p-\hat p}{\sqrt{\frac{\hat p(1-\hat p)}{n}}}>\frac{0.46-0.25}{\sqrt{\frac{0.25(1-0.25)}{747}}})

                   =P(Z>13.25)\\=1-P(Z

 *Use a <em>z</em> table for the probability.

The probability value is almost equal to 0. This probability is very low indicating that the proportion of people dying  in that particular interval if deaths occurred randomly throughout the year is unusual.

3 0
3 years ago
Write an equation in point-slope form of the line through point J(4,1) with slope -4.
Nimfa-mama [501]
Point slope form
y - y1 = m(x - x1)

Plug in the given
slope = -4
Point ( 4, 1)

y - 1 = -4(x - 4)
y - 1 = -4x +16
add 1 to both sides
y = -4x + 17

Hope this helps!
4 0
3 years ago
Read 2 more answers
2x+3y=12 how do I solve for x and y
Elena-2011 [213]

2x+3y= 12

To solve for the x intercept, y=0

2x+3(0)= 12

2x= 12

Divide both sides by 2

2x/2x= 12/2x

X= 6

So:(6,0)

To solve for y intercept, x=0

2(0)+3y= 12

3y= 12

Divide both sides by 3

3y/3y= 12/3

Y= 4

So: (0,4)


4 0
3 years ago
Read 2 more answers
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