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Licemer1 [7]
3 years ago
10

A herd of dinosaurs made paintings in the sand with their claws. Each baby dinosaur made 15 paintings and each adult dinosaur ma

de 7 paintings. The entire herd made 208 paintings in total, and there were 3 times as many baby dinosaurs. How many baby dinosaurs and adult dinosaurs were there?
Mathematics
2 answers:
weeeeeb [17]3 years ago
6 0

12 babies 4 adults

Step-by-step explanation:

tia_tia [17]3 years ago
5 0
Let x = baby dinosaurs
     y = adult dinosaurs
equation 1
total paintings = 15 painting x + 7 paintings y
equation 2
baby dinosaurs = 3 times adult
x= 3y
substitute x= 3y to equation 1
y = 4 
x = 12
There are 12 baby dinosaurs and 4 adult dinosaurs in a herd with 208 total paintings 


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The sum of two numbers is 36 . The larger number is 22 more than the smaller number. What are the numbers?
Setler [38]
X + y = 36
x = y + 22

y + 22 + y = 36
2y + 22 = 36
2y = 36 - 22
2y = 14
y = 14/2
y = 7

x + y = 36
x + 7 = 36
x = 36 - 7
x = 29

so ur 2 numbers are : 7 and 29
4 0
3 years ago
A cola-dispensing machine is set to dispense 8 ounces of cola per cup, with a standard deviation of 1.0 ounce. The manufacturer
pshichka [43]

Answer:

Step-by-step explanation:

Hello!

The variable of interest is X: ounces per cup dispensed by the cola-dispensing machine.

The population mean is known to be μ= 8 ounces and its standard deviation σ= 1.0 ounce. Assuming the variable has a normal distribution.

A sample of 34 cups was taken:

a. You need to calculate the Z-values corresponding to the top 5% of the distribution and the lower 5% of it. This means you have to look for both Z-values that separates two tails of 5% each from the body of the distribution:

The lower value will be:

Z_{o.o5}= -1.648

You reverse the standardization using the formula Z= \frac{X[bar]-Mu}{\frac{Sigma}{\sqrt{n} } } ~N(0;1)

-1.648= \frac{X[bar]-8}{\frac{1}{\sqrt{34} } }

X[bar]= 7.72ounces

The lower control point will be 7.72 ounces.

The upper value will be:

Z_{0.95}= 1.648

1.648= \frac{X[bar]-8}{\frac{1}{\sqrt{34} } }

X[bar]= 8.28ounces

The upper control point will be 8.82 ounces.

b. Now μ= 7.6, considering the control limits of a.

P(7.72≤X[bar]≤8.28)= P(X[bar]≤8.28)- P(X[bar]≤7.72)

P(Z≤(8.28-7.6)/(1/√34))- P(Z≤7.72-7.6)/(1/√34))

P(Z≤7.11)- P(Z≤0.70)= 1 - 0.758= 0.242

There is a 0.242 probability of the sample means being between the control limits, this means that they will be outside the limits with a probability of 1 - 0.242= 0.758, meaning that the probability of the change of population mean being detected is 0.758.

b. For this item μ= 8.7, the control limits do not change:

P(7.72≤X[bar]≤8.28)= P(X[bar]≤8.28)- P(X[bar]≤7.72)

P(Z≤(8.28-8.7)/(1/√34))- P(Z≤7.72-8.7)/(1/√34))

P(Z≤-2.45)- P(Z≤-5.71)=0.007 - 0= 0.007

There is a 0.007 probability of not detecting the mean change, which means that you can detect it with a probability of 0.993.

I hope it helps!

5 0
3 years ago
Read 2 more answers
What is the area of a circle if the circumference is 29
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Answer:

67 square units

Step-by-step explanation:

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8 0
2 years ago
In a study of government financial aid for college​ students, it becomes necessary to estimate the percentage of​ full-time coll
ICE Princess25 [194]

Answer:

a) n = 1037.

b) n = 1026.

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of \pi, and a confidence level of 1-\alpha, we have the following confidence interval of proportions.

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

In which

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The margin of error is:

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

99% confidence level

So \alpha = 0.01, z is the value of Z that has a pvalue of 1 - \frac{0.01}{2} = 0.995, so Z = 2.575.

​(a) Assume that nothing is known about the percentage to be estimated.

We need to find n when M = 0.04.

We dont know the percentage to be estimated, so we use \pi = 0.5, which is when we are going to need the largest sample size.

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

0.04 = 2.575\sqrt{\frac{0.5*0.5}{n}}

0.04\sqrt{n} = 2.575*0.5

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n = 1037.

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\pi = 0.55

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

0.04 = 2.575\sqrt{\frac{0.55*0.45}{n}}

0.04\sqrt{n} = 2.575*\sqrt{0.55*0.45}

(\sqrt{n}) = \frac{2.575*\sqrt{0.55*0.45}}{0.04}

(\sqrt{n})^{2} = (\frac{2.575*\sqrt{0.55*0.45}}{0.04})^{2}

n = 1025.7

Rounding up

n = 1026.

6 0
3 years ago
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6 0
4 years ago
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