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12345 [234]
3 years ago
14

The temperature was recorded for four straight days.The temperatures was recorded for the first three days were -2°,10°,and 7°.W

hat must be the temperature recorded on the fourth day if the total temperature change was -4°?
Mathematics
1 answer:
lubasha [3.4K]3 years ago
7 0

Answer:

-19^{\circ}

Step-by-step explanation:

Let x represent temperature on 4th day.

We have been given that the temperatures was recorded for the first three days were -2°,10°,and 7°.

The total temperature on f days would be -2^{\circ}+10^{\circ}+7^{\circ}+x.

We have been given that total temperature change was -4°. We can represent this information in an equation as:

-2^{\circ}+10^{\circ}+7^{\circ}+x=-4^{\circ}

15^{\circ}+x=-4^{\circ}

x=-4^{\circ}-15^{\circ}

x=-19^{\circ}

Therefore, the temperature on 4th day must be equal to -19^{\circ}.

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2. Suppose that a particular medical procedure has a cost that is approximately normally distributed with a mean of $19,800 and
Lina20 [59]

Answer:

a) 0.5085

b) -1.6551

c) 0.8103834

Step-by-step explanation:

Data provided in the question:

Mean = $19,800

Standard deviation, s = $2,900

Now,

z score = [ X - Mean ] ÷ s

a) The procedure costs between $18,000 and $22,000

For X = $22,000

z-score = [ $22,000 - $19,800 ] ÷ $2,900

= 0.7586

For X = $18,000

z-score = [ $18,000 - $19,800 ] ÷ $2,900

= -0.62069

P(procedure costs between $18,000 and $22,000)

= P(z < 0.7586) - P( z < -0.62069)

= 0.7759541 - 0.2674018           [ P value from standard z table ]

= 0.5085

b) The procedure costs less than $15,000

For X = $15,000

z-score = [ $15,000 - $19,800 ] ÷ $2,900

= -1.6551

thus,

The procedure costs less than $15,000

P (z < -1.655172 )

= 0.0489448                      [ P value from standard z table ]

c) The procedure costs more than $17,250

z-score = [ $17,250 - $19,800 ] ÷ $2,900

= -0.87931

thus,

The procedure costs more than $17,250.

P (z > -0.87931  ) = 0.8103834

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