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monitta
3 years ago
5

Write an equation for the following situation:

Mathematics
1 answer:
MariettaO [177]3 years ago
7 0
Let
x---------> <span>represent the number of weeks
y---------> </span><span>represent the amount of money saved

we know that
for x=6 weeks    y=40*6-------> $240
then 
300-240--------> $60----------> initial amount for x=0

the equation is 
y=40*x+60

the answer is
</span>y=40*x+60<span>

</span>
You might be interested in
Historically, a certain region has experienced 65 thunder days annually. (A "thunder day" is day on which at least one instance
Makovka662 [10]

Answer:

No, we can't conclude that the mean number of thunder days is less than 65.

Step-by-step explanation:

We are given that Historically, a certain region has experienced 65 thunder days annually. Over the past eleven years, the mean number of thunder days is 55 with a standard deviation of 20.

We have to test that the mean number of thunder days is less than 65 or not.

Let, NULL HYPOTHESIS, H_0 : \mu \geq 65 days {means that the mean number of thunder days is higher than or equal to 65}

ALTERNATE HYPOTHESIS, H_1 : \mu < 65 days {means that the mean number of thunder days is less than 65}

The test statistics that will be used here is One-sample t-test;

               T.S. = \frac{\bar X - \mu}{\frac{s}{\sqrt{n} } } ~ t_n_-_1

where,  \bar X = sample mean number of thunder days = 55

              s = sample standard deviation = 20

              n = number of years = 11

So, <u>test statistics </u>= \frac{55-65}{\frac{20}{\sqrt{11} } }  ~ t_1_0

                             = -1.658

Now, at 1% significance level t table gives critical value of -2.764. Since our test statistics is higher than the critical value of t so we have insufficient evidence to reject null hypothesis as it will not fall in the left side of critical value rejection region.

Therefore, we conclude that the mean number of thunder days is higher than or equal to 65.

6 0
3 years ago
How do you find the answer for this Solve for Y -y= -33
Degger [83]

Answer:

the equation is

Y-y = -33

-33y

3 0
3 years ago
The time that passes between the time you see lightning and you hear the thunder depends on your distance from the lightning. Wi
xxTIMURxx [149]

When the lightning occurs 4.5 km away then we will hear after 13.5 seconds and if we hear after 1.5 seconds then we are 0.5 km away from lightning.

Given With each km from the lightning 3 seconds passes.

a) The table of values for distances from 0 to 5 km.

     Distances                     Time

            0                              0

            1                               3

            2                              6

            3                               9

            4                               12

            5                                15

b) The graph of lightning is a linear relation because it is increasing constantly by 3 seconds. f(x)=3x

c) When distance =1  , the time be 3 seconds

when distance =4.5  , the time=3*4.5

=13.5 seconds.

d) When time is 3 seconds = 1km

when time =1.5 seconds , the distance be 1/2=0.5 km.

Learn more about graph at brainly.com/question/4025726

#SPJ10

5 0
2 years ago
£2500 is invested at 3.6% compound interest per annum.
kirill115 [55]

Answer:

its going to be 20 years

Step-by-step explanation:

1.036

4 0
3 years ago
GEOMETRY
ira [324]

Answer:

Its like how much you multiply the original shape by, to the new one.

For example, if your new shape is double the size of the original shape, then the scale factor is 2, because you would multiply the original by 2, to get the size of the new.

However, if your new shape is half the size, then the scale factor is 1/2, as you multiply the original by half to get the new shape.

Hope this cleared things up, lmk down below if you any further questions.

8 0
3 years ago
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