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Xelga [282]
2 years ago
11

A chef at a restaurant cooks 90 meals in one day. 35% of the 90 meals were vegetarian. How many meals were vegetarian?​

Mathematics
1 answer:
serg [7]2 years ago
6 0
Answer is 31.5% vegetarian. Here is the my work. Hope this helps!

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What is the mean of the following numbers: 5, 7, 4, 4, 5, 4, 6?
jeka57 [31]

Answer:

5

Step-by-step explanation:

5+7+4+4+5+4+6 = 35

35 / 7 = 5

8 0
4 years ago
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The angle of sector of cicle of diameter 8cm is 135 . find the area of the sector<br>​
Naddik [55]

Answer:

18.85 cm²

Step-by-step explanation:

The formula to find the area of a sector = θ/360 × πr²

We are given:

θ = Angle of the sector = 135°

Radius = Diameter/2

Diameter = 8cm , Hence Radius = 8cm/2 = 4cm

Hence,

Area of a sector = 135/360 × π ×(4cm)²

= 18.849555922 cm²

Approximately to the nearest hundredth = 18.85 cm²

Therefore, the area of the sector is 18.85 cm²

8 0
3 years ago
A couple decide to have 5 children what if the probability that they will have at least one girl
Alchen [17]
The chance that they will have one girl is 1:5 because they are having 5 children and one is going to be a girl
8 0
3 years ago
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PLEASE HURRY IF YOU ANSWER AND IT'S RIGHT YOU GET BRAINLYEST
mezya [45]
The answer is 130cm
5 0
3 years ago
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From a large number of actuarial exam scores, a random sample of scores is selected, and it is found that of these are passing s
Mnenie [13.5K]

<u>Supposing 60 out of 100 scores are passing scores</u>, the 95% confidence interval for the proportion of all scores that are passing is (0.5, 0.7).

  • The lower limit is 0.5.
  • The upper limit is 0.7.

In a sample with a number n of people surveyed with a probability of a success of \pi, and a confidence level of \alpha, we have the following confidence interval of proportions.

\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}

In which

z is the z-score that has a p-value of \frac{1+\alpha}{2}.

60 out of 100 scores are passing scores, hence n = 100, \pi = \frac{60}{100} = 0.6

95% confidence level

So \alpha = 0.95, z is the value of Z that has a p-value of \frac{1+0.95}{2} = 0.975, so z = 1.96.

The lower limit of this interval is:

\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.6 - 1.96\sqrt{\frac{0.6(0.4)}{100}} = 0.5

The upper limit of this interval is:

\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.6 + 1.96\sqrt{\frac{0.6(0.4)}{100}} = 0.7

The 95% confidence interval for the proportion of all scores that are passing is (0.5, 0.7).

  • The lower limit is 0.5.
  • The upper limit is 0.7.

A similar problem is given at brainly.com/question/16807970

5 0
3 years ago
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