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4 years ago

URGENT!!! PLEASE LOOK AT THE PICTURE!!! PLEASE HELP!!!!

Mathematics

1 answer:

Svetllana [295]4 years ago

7 0

Answer:

Step-by-step explanation:

this is a 45-45-90 triangle. The angle opposite the 45° angle is also 45°.

HI = HJ because this is an isosceles triangle.

Use the cos function to find the length of HJ:

cos Ф = adj / hyp or adj = (hyp)(cos Ф).

Here we have adj = HI = HJ = 4√5 cos 45° = 4√5 · -----

√2, or 4√5 / √2

4√10

This last result simplifies to ----------

or 2√10 (answer)

The length of side HJ and that of HI is 2/√10.

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A random sample of 10 parking meters in a resort community showed the following incomes for a day. Assume the incomes are normal

GenaCL600 [577]

Answer:

A 95% confidence interval for the true mean is [$3.39, $6.01].

Step-by-step explanation:

We are given that a random sample of 10 parking meters in a resort community showed the following incomes for a day;

Incomes (X): $3.60, $4.50, $2.80, $6.30, $2.60, $5.20, $6.75, $4.25, $8.00, $3.00.

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

P.Q. = $\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }$ ~ $t_n_-_1$

where, $\bar X$ = sample mean income = $\frac{\sum X}{n}$ = $4.70

s = sample standard deviation = $\sqrt{\frac{\sum (X-\bar X)^{2} }{n-1} }$ = $1.83

n = sample of parking meters = 10

$\mu$ = population mean

Here for constructing a 95% confidence interval we have used a One-sample t-test statistics because we don't know about population standard deviation.

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

P(-2.262 < $t_9$ < 2.262) = 0.95 {As the critical value of t at 9 degrees of

freedom are -2.262 & 2.262 with P = 2.5%}

P(-2.262 < $\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }$ < 2.262) = 0.95

P( $-2.262 \times {\frac{s}{\sqrt{n} } }$ < ${\bar X-\mu$ < $2.262 \times {\frac{s}{\sqrt{n} } }$ ) = 0.95

P( $\bar X-2.262 \times {\frac{s}{\sqrt{n} } }$ < $\mu$ < $\bar X+2.262 \times {\frac{s}{\sqrt{n} } }$ ) = 0.95

95% confidence interval for $\mu$ = [ $\bar X-2.262 \times {\frac{s}{\sqrt{n} } }$ , $\bar X+2.262 \times {\frac{s}{\sqrt{n} } }$ ]

= [ $4.70-2.262 \times {\frac{1.83}{\sqrt{10} } }$ , $4.70+ 2.262 \times {\frac{1.83}{\sqrt{10} } }$ ]

= [$3.39, $6.01]

Therefore, a 95% confidence interval for the true mean is [$3.39, $6.01].

The interpretation of the above result is that we are 95% confident that the true mean will lie between incomes of $3.39 and $6.01.

Also, the margin of error = $2.262 \times {\frac{s}{\sqrt{n} } }$

= $2.262 \times {\frac{1.83}{\sqrt{10} } }$ = 1.31

4 0

3 years ago

The data set represents a progression of hourly temperature measurements.Use the regression equation to predict the temperature

zimovet [89]

I will attach google sheet that I used to find regression equation.
We can see that linear fit does work, but the polynomial fit is much better.
We can see that R squared for polynomial fit is higher than R squared for the linear fit. This tells us that polynomials fit approximates our dataset better.
This is the polynomial fit equation:
$T(h)=20.2-3.6h-0.875h^2$
I used h to denote hours. Our prediction of temperature for the sixth hour would be:
$T(6)=20.2-3.6(6)-0.875(6)^2=-32.9$
Here is a link to the spreadsheet (https://docs.google.com/spreadsheets/d/17awPz5U8Kr-ZnAAtastV-bnvoKG5zZyL3rRFC9JqVjM/edit?usp=sharing)