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ahrayia [7]
3 years ago
9

Plz help! Giving away more points than usual!!! Thxs :) (btw the rate of change isn't 8 nor 29, I tried them both like 500 times

)

Mathematics
2 answers:
o-na [289]3 years ago
6 0

Answer:

rate change: -8

Equation: y = 232 -8x

After 5 hours: 192

Step-by-step explanation:

8 are destroyed every hour, so there are 8 less every hour, so the rate of change is -8

It starts with 232 and loses 8 per hour, so the formula is

y = 232 - 8x

After 5 hours:

y = 232 - 8(5)

    y = 232 - 40

         y = 192

mestny [16]3 years ago
5 0

Answer:

Step-by-step explanation:

The rate of change in this case is -8, since the amount is decreasing.

The equation is:

y=232-8x.

After 5 hours, he will have 192 rews.

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Find the slop for the given graph and y-intercept <br><br> I got a pic to lol
Sergeeva-Olga [200]
The slope is rise/run
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3 0
3 years ago
Tickets for Adults were $14.50 and tickets for Children were $9.00. We purchased 12 tickets and spent a total of $135.50 to get
mihalych1998 [28]

Answer:

(5 , 7)

Step-by-step explanation:

let a = the number of adult tickets

let c = the number of children tickets

then we have to solve the following system

a + c = 12

14.5a + 9c = 135.5

we proceed by Substitution:

              c = 12 - a

14.5a + 9c = 135.5

After some calculations we find

a=5 and c=7

6 0
3 years ago
Read 2 more answers
The tensile strength of stainless steel produced by a plant has been stable for a long time with a mean of 72 kg/mm2 and a stand
Elanso [62]

Answer:

95% confidence interval for the mean of tensile strength after the machine was adjusted is [73.68 kg/mm2 , 74.88 kg/mm2].

Yes, this data suggest that the tensile strength was changed after the adjustment.

Step-by-step explanation:

We are given that the tensile strength of stainless steel produced by a plant has been stable for a long time with a mean of 72 kg/mm 2 and a standard deviation of 2.15.

A machine was recently adjusted and a sample of 50 items were taken to determine if the mean tensile strength has changed. The mean of this sample is 74.28. Assume that the standard deviation did not change because of the adjustment to the machine.

Firstly, the pivotal quantity for 95% confidence interval for the population mean is given by;

                         P.Q. = \frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }  ~ N(0,1)

where, \bar X = sample mean strength of 50 items = 74.28

            \sigma = population standard deviation = 2.15

            n = sample of items = 50

            \mu = population mean tensile strength after machine was adjusted

<em>Here for constructing 95% confidence interval we have used One-sample z test statistics as we know about population standard deviation.</em>

So, 95% confidence interval for the population mean, \mu is ;

P(-1.96 < N(0,1) < 1.96) = 0.95  {As the critical value of z at 2.5% level of

                                                  significance are -1.96 & 1.96}  

P(-1.96 < \frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } } < 1.96) = 0.95

P( -1.96 \times {\frac{\sigma}{\sqrt{n} } } < {\bar X-\mu} < 1.96 \times {\frac{\sigma}{\sqrt{n} } } ) = 0.95

P( \bar X-1.96 \times {\frac{\sigma}{\sqrt{n} } } < \mu < \bar X+1.96 \times {\frac{\sigma}{\sqrt{n} } } ) = 0.95

<u><em>95% confidence interval for</em></u> \mu = [ \bar X-1.96 \times {\frac{\sigma}{\sqrt{n} } } , \bar X+1.96 \times {\frac{\sigma}{\sqrt{n} } } ]

                 = [ 74.28-1.96 \times {\frac{2.15}{\sqrt{50} } } , 74.28+1.96 \times {\frac{2.15}{\sqrt{50} } } ]

                 = [73.68 kg/mm2 , 74.88 kg/mm2]

Therefore, 95% confidence interval for the mean of tensile strength after the machine was adjusted is [73.68 kg/mm2 , 74.88 kg/mm2].

<em>Yes, this data suggest that the tensile strength was changed after the adjustment as earlier the mean tensile strength was 72 kg/mm2 and now the mean strength lies between 73.68 kg/mm2 and 74.88 kg/mm2 after adjustment.</em>

8 0
3 years ago
Suppose your​ friend's parents invest 15,000 in an account paying 7% compounded annually. What will the balance be after 5 ​year
meriva

Answer:

FV= $21,038.28

Step-by-step explanation:

Giving the following information:

Initial investment (PV)= $15,000

Interest rate (i)= 7% compounded annually

Number of periods (n)= 5

<u>To calculate the future value (FV), we need to use the following formula:</u>

FV= PV*(1 + i)^n

FV= 15,000*(1.07^5)

FV= $21,038.28

6 0
2 years ago
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