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3 years ago

7

The temperature on sunday was -15 c.The temperature on monday was 12 degrees less than the temperature on sunday . What was the

temperature on monday

2 answers:

Anon25 [30]3 years ago

5 0

Temperature on Sunday = -15℃

The temperature on monday was 12 degrees less than the temperature on sunday

Temperature on Monday = -15℃ - 12℃ = -27℃

Temperature on Monday is -27℃.

FrozenT [24]3 years ago

4 0

Temperature on sunday was-15c

Temperature on monday was 12c less than sunday temperature

So temperature on monday=-15-12

=-27c

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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

<em>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.</em>

<u>So, 95% confidence interval for the population mean, </u> $\mu$ <u> is ;</u>

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

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

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

= [$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 = $2.262 \times {\frac{s}{\sqrt{n} } }$

= $2.262 \times {\frac{1.83}{\sqrt{10} } }$ = <u>1.31</u>

4 0

3 years ago

Fill in the missing numbers to create equivalent ratios.

jasenka [17]

Step-by-step explanation:

The way to find missing numbers in equivalent ratios is to multiply the means (the first denominator and the second numerator) and multiply the extremes (the first numerator and the second denominator). It sounds really complicated, but it is quite simple =)

2/5 = x/10

The means in this equivalent ration are 5 and x. The extremes are 2 and 10.

5x = 20

Now solve =)

x = 4

That was pretty simple. Let's move on to the next one. Do exactly the same thing here:

4/10 = 6/x

60 = 4x

15 = x

That was pretty simple, too! Keep going!

6/15 = x/25

15x = 150

x = 10

All of these should be equal, so check them by dividing:

2/5 = 0.4

4/10 = 0.4

6/15 = 0.4

10/25 = 0.4

They all check out, so these are your answers: 2/5, 4/10, 6/15, 10/25

I really hope this helps you =)

8 0

3 years ago

Read 2 more answers

Yu Xing paid 3.60 dollar for 2 pens after a 10 percent discount .What was the usual price of 1 pen.

notka56 [123]

Answer:

2 $

Step-by-step explanation:

let the original price be x

price after 10% discount = 3.60$

$x - \frac{10}{100} \times x = 3.60$

$\frac{100x - 10x}{100} = 3.60 \\ \frac{90x}{100} = 3.60 \\ 9x = 36 \\ x = 4$

The original price of 2 pens is 4$

original price of one pen = 4/ 2

<u>= 2 $</u>

8 0

2 years ago

Determine the sum of: −10y2+(−3y2)−4y2−(−6y2)<br><br> the 2 are squared

Bas_tet [7]

-10y^2 + (-3y^2) - 4y^2 - (-6y^2) =
-10y^2 - 3y^2 - 4y^2 + 6y^2 =
-17y^2 + 6y^2 =
- 11y^2

4 0

3 years ago

A person purchased a slot machine and tested it by playing it 1,137 times. There are 10 different categories of outcomes, includ

ivann1987 [24]

Answer:

$p_v= P(\chi^2_{9}>11.517)=0.2419$

And on this case if we see the significance level given $\alpha=0.1$ we see that $p_v>alpha$ so we fail to reject the null hypothesis that the observed outcomes agree with the expected frequencies at 10% of significance.

Step-by-step explanation:

A chi-square goodness of fit test determines if a sample data obtained fit to a specified population.

$p_v$ represent the p value for the test

O= obserbed values

E= expected values

The system of hypothesis for this case are:

Null hypothesis: $O_i = E_i[/tex[Alternative hypothesis: [tex]O_i \neq E_i$

The statistic to check the hypothesis is given by:

$\chi^2 =\sum_{i=1}^n \frac{(O_i -E_i)^2}{E_i}$

On this case after calculate the statistic they got: $\chi^2 = 11.517$

And in order to calculate the p value we need to find first the degrees of freedom given by:

$df=n-1=10-1=9$ , where k represent the number of levels (on this cas we have 10 categories)

And in order to calculate the p value we need to calculate the following probability:

$p_v= P(\chi^2_{9}>11.517)=0.2419$

And on this case if we see the significance level given $\alpha=0.1$ we see that $p_v>alpha$ so we fail to reject the null hypothesis that the observed outcomes agree with the expected frequencies at 10% of significance.

6 0

3 years ago