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irinina [24]
3 years ago
15

in the town zoo 3/28 of the animals are birds . of the birds, 4/15 are birds of prey . what fraction of.the animal at zoo are bi

rds of prey
Mathematics
1 answer:
salantis [7]3 years ago
3 0
The Free Application for Federal Student Aid (FAFSA) is a form that can be prepared annually by current and prospective college students (undergraduate and graduate) in the United States to determine their eligibility for student financial aid.
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What is 18-0.9 if you know answer pls lol
Harlamova29_29 [7]

Answer:

17.1

Step-by-step explanation:

3 0
3 years ago
Solve the equation<br> 7h-5(3h-8) = 72
frosja888 [35]

Answer:

h = -4

Step-by-step explanation:

7h-5(3h-8) = 72

Distribute

7h - 15h +40 = 72

Combine like terms

-8h +40 = 72

Subtract 40 from each side

-8h+40-40 = 72-40

-8h = 32

Divide each side by -8

-8h/-8 = -32/-8

h = -4

3 0
3 years ago
Read 2 more answers
Given that P=x+y. Find P when: x=−5 and y=−3
Sergio [31]

Answer:

P = - 8

Step-by-step explanation:

Given

P = x + y ← substitute x = - 5 and y = - 3

P = - 5 + (- 3) = - 5 - 3 = - 8

8 0
3 years ago
Boxes of raisins are labeled as containing 22 ounces. Following are the weights, in the ounces, of a sample of 12 boxes. It is r
ZanzabumX [31]

Answer:

A 90% confidence interval for the mean weight is [21.78 ounces, 21.98 ounces].

Step-by-step explanation:

We are given the weights, in the ounces, of a sample of 12 boxes below;

Weights (X): 21.88, 21.76, 22.14, 21.63, 21.81, 22.12, 21.97, 21.57, 21.75, 21.96, 22.20, 21.80.

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 weight = \frac{\sum X}{n} = 21.88 ounces

            s = sample standard deviation = \sqrt{\frac{\sum (X-\bar X)^{2} }{n-1} }  = 0.201 ounces

            n = sample of boxes = 12

            \mu = population mean weight

<em>Here for constructing a 90% confidence interval we have used a One-sample t-test statistics because we don't know about population standard deviation.</em>

<u>So, 90% confidence interval for the population mean, </u>\mu<u> is ;</u>

P(-1.796 < t_1_1 < 1.796) = 0.90  {As the critical value of t at 11 degrees of

                                                  freedom are -1.796 & 1.796 with P = 5%}  

P(-1.796 < \frac{\bar X-\mu}{\frac{s}{\sqrt{n} } } < 1.796) = 0.90

P( -1.796 \times {\frac{s}{\sqrt{n} } } < {\bar X-\mu} < 1.796 \times {\frac{s}{\sqrt{n} } } ) = 0.90

P( \bar X-1.796 \times {\frac{s}{\sqrt{n} } } < \mu < \bar X+1.796 \times {\frac{s}{\sqrt{n} } } ) = 0.90

<u>90% confidence interval for</u> \mu = [ \bar X-1.796 \times {\frac{s}{\sqrt{n} } } , \bar X+1.796 \times {\frac{s}{\sqrt{n} } } ]

                                        = [ 21.88-1.796 \times {\frac{0.201}{\sqrt{12} } } , 21.88+1.796 \times {\frac{0.201}{\sqrt{12} } } ]

                                        = [21.78, 21.98]

Therefore, a 90% confidence interval for the mean weight is [21.78 ounces, 21.98 ounces].

8 0
3 years ago
Marianna's mom wants to lose at least 10 pounds. She created a coordinate grid to
goldenfox [79]

Answer:

week 12

Step-by-step explanation:

not sure if i can read it correctly but it looks like she gets below 120 after 12 lines. plus 130-12 is less than 120

6 0
3 years ago
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