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Lapatulllka [165]
3 years ago
9

Arianna went to the store to buy some chicken. The prince per pound of the chicken is $5.75 per pound and she has a coupon for $

3 off the final amount. With the coupon how much would arianna have to pay to buy 4 pounds of chicken
Mathematics
2 answers:
gregori [183]3 years ago
6 0

Answer:

$20

Step-by-step explanation:

If you buy 4 pounds of chicken, you multiply 5.75 times 4. That gives you $23. You have a coupon for $3 off the final amount. $23-$3 = $20.

Ray Of Light [21]3 years ago
5 0

Answer:

20

Step-by-step explanation:

(5.75x4)-3

23-3

20

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What is 53 divided by 76
Goshia [24]

Answer:

0.69

Step-by-step explanation:

3 0
3 years ago
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The temperature of a certain solution is estimated by taking a large number of independent measurements and averaging them. The
Mademuasel [1]

Answer:

(a) The 95% confidence interval for the temperature is (36.80°C, 37.20°C).

(b) The confidence level is 68%.

(c) The necessary assumption is that the population is normally distributed.

(d) The 95% confidence interval for the temperature if 10 measurements were made is (36.93°C, 37.07°C).

Step-by-step explanation:

Let <em>X</em> = temperature of a certain solution.

The estimated mean temperature is, \bar x=37^{o}C.

The estimated standard deviation is, s=0.1^{o}C.

(a)

The general form of a (1 - <em>α</em>)% confidence interval is:

CI=SS\pm CV\times SD

Here,

SS = sample statistic

CV = critical value

SD = standard deviation

It is provided that a large number of independent measurements are taken to estimate the mean and standard deviation.

Since the sample size is large use a <em>z</em>-confidence interval.

The critical value of <em>z</em> for 95% confidence interval is:

z_{0.025}=1.96

Compute the confidence interval as follows:

CI=SS\pm CV\times SD\\=37\pm 1.96\times 0.1\\=37\pm0.196\\=(36.804, 37.196)\\\approx (36.80^{o}C, 37.20^{o}C)

Thus, the 95% confidence interval for the temperature is (36.80°C, 37.20°C).

(b)

The confidence interval is, 37 ± 0.1°C.

Comparing the confidence interval with the general form:

37\pm 0.1=SS\pm CV\times SD

The critical value is,

CV = 1

Compute the value of P (-1 < Z < 1) as follows:

P(-1

The percentage of <em>z</em>-distribution between -1 and 1 is, 68%.

Thus, the confidence level is 68%.

(c)

The confidence interval for population mean can be constructed using either the <em>z</em>-interval or <em>t</em>-interval.

If the sample selected is small and the standard deviation is estimated from the sample, then a <em>t</em>-interval will be used to construct the confidence interval.

But this will be possible only if we assume that the population from which the sample is selected is Normally distributed.

Thus, the necessary assumption is that the population is normally distributed.

(d)

For <em>n</em> = 10 compute a 95% confidence interval for the temperature as follows:

The (1 - <em>α</em>)% <em>t</em>-confidence interval is:

CI=\bar x\pm t_{\alpha/2, (n-1)}\times \frac{s}{\sqrt{n}}

The critical value of <em>t</em> is:

t_{\alpha/2, (n-1)}=t_{0.025, 9}=2.262

*Use a <em>t</em>-table for the critical value.

The 95% confidence interval is:

CI=37\pm 2.262\times \frac{0.1}{\sqrt{10}}\\=37\pm 0.072\\=(36.928, 37.072)\\\approx (36.93^{o}C, 37.07^{o}C)

Thus, the 95% confidence interval for the temperature if 10 measurements were made is (36.93°C, 37.07°C).

3 0
3 years ago
I need help asap.
DaniilM [7]
1+10 < or equal to 40
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Helppppppppopp asappppppppp
serg [7]

Answer:

10 = 74°

7 is alternate to 12 so 12 is 74°

4 0
3 years ago
HELP WITH #2 PLEASEEE
Shtirlitz [24]
ILL HELP IF YOU HELP ME I NEED HELP ALSO
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3 years ago
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