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Schach [20]
3 years ago
14

A cake is made from the ingredients listed below.

Mathematics
1 answer:
Oksana_A [137]3 years ago
5 0

To make a cake :-

Quantity of flour needed = 500 g

Quantity of butter needed = 450 g

Quantity of sugar needed = 470 g

Quantity of mixed fruit needed = 1.8 kg = 1000 g + 800 g

Quantity of eggs needed = 70 × 4 = 280 g

It's weight after adding all the ingredients =

=500 + 450 + 470 + 1000 + 800 + 280

= 3500 g

Mass that got lost while baking the cake :

<h3>= 12 % of 3500 </h3>

= \frac{12}{100}  \: of \: 3500

=  \frac{12}{100}  \times 3500

=\frac{(12 \times 3500)}{100}

=\frac{42000}{100}  \: g

= 420 \: g

Final mass = Initial mass - mass that got lost

= 3500 - 420 \: g

= 3080 \: g

∴ The final mass of the cake is 3080 g .

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lana [24]

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4 0
3 years ago
Data were collected on random specimens of Yellowfin tuna from 1991 and 2010. For the sample of 231 specimens of Yellowfin tuna,
Molodets [167]

Answer:

95% confidence interval: (0.325 ,0.383)

Step-by-step explanation:

We are given the following in the question:

Sample size, n = 231

Sample mean = 0.354 ppm

Sample standard deviation = 0.231 ppm

95% confidence interval:

\bar{x} \pm t_{critical}\displaystyle\frac{s}{\sqrt{n}}  

Putting the values, we get,  

t_{critical}\text{ at degree of freedom 4 and}~\alpha_{0.01} = \pm 1.97  

0.354 \pm 1.97(\dfrac{0.231}{\sqrt{231}} ) = 0.354 \pm 0.029 = (0.325 ,0.383)  

6 0
3 years ago
A healthcare provider monitors the number of CAT scans performed each month in each of its clinics. The most recent year of data
Rasek [7]

Answer: a. 1.981 < μ < 2.18

              b. Yes.

Step-by-step explanation:

A. For this sample, we will use t-distribution because we're estimating the standard deviation, i.e., we are calculating the standard deviation, and the sample is small, n = 12.

First, we calculate mean of the sample:

\overline{x}=\frac{\Sigma x}{n}

\overline{x}=\frac{2.31=2.09+...+1.97+2.02}{12}

\overline{x}= 2.08

Now, we estimate standard deviation:

s=\sqrt{\frac{\Sigma (x-\overline{x})^{2}}{n-1} }

s=\sqrt{\frac{(2.31-2.08)^{2}+...+(2.02-2.08)^{2}}{11} }

s = 0.1564

For t-score, we need to determine degree of freedom and \frac{\alpha}{2}:

df = 12 - 1

df = 11

\alpha = 1 - 0.95

α = 0.05

\frac{\alpha}{2}= 0.025

Then, t-score is

t_{11,0.025} = 2.201

The interval will be

\overline{x} ± t.\frac{s}{\sqrt{n} }

2.08 ± 2.201\frac{0.1564}{\sqrt{12} }

2.08 ± 0.099

The 95% two-sided CI on the mean is 1.981 < μ < 2.18.

B. We are 95% confident that the true population mean for this clinic is between 1.981 and 2.18. Since the mean number performed by all clinics has been 1.95, and this mean is less than the interval, there is evidence that this particular clinic performs more scans than the overall system average.

8 0
3 years ago
A science fair poster is very rectangular 3 feet long and 2 feet wide what is the area of the poster in square inches
pantera1 [17]
You want to do base × height to get your answer
3×2 = 6
So the area of the poster is 6 sq in.


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3 years ago
A school play earned over $250 from selling tickets, which cost $11 each. Which of the following inequalities models this situat
jeyben [28]
Its B because its supposed to be OVER 250
7 0
3 years ago
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