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2 years ago

Answer me pls I will mark brainliest worth 50 points. (sorry when I put 100 it goes to 50).

Mathematics

2 answers:

Anettt [7]2 years ago

6 0

Answer:

■ A random sample of dogs at different animal shelters in Abu Dhabi shows that 12 of the 60 dogs are puppies. Abu Dhabi's animal shelters collectively house 1,200 dogs each year. About how many dogs in all of the city’s animal shelters are puppies?

➢12/60 are puppies. 60 are the number of dogs, 12 is number of puppies.

If there are 1200 dogs, we focus on the 60. 1200/60= 20

Whatever we do to the denominator we do to the numerator, so since we can multiply 60 by 20 to get 1200; we need to multiply 12 by 20 to get the total number of dogs. Which is 240.

60/1200 = x/20 1200x /1200 =12/20 x =240

$TextFromYourMrEx$

#CarryOnLearning

mars1129 [50]2 years ago

4 0

Answer:

240

Step-by-step explanation:

1) First you set up a proportion.

2) You do 12/60=x/1,200.

3) You cross multiply and do 12x1,200. Your answer is 14400.

4) Then you do 14400/60=240

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The nth term of the GP is given by

givi [52]

Answer:

Step-by-step explanation:

hello :

An = 10 -10/2^n

the first 4 terms are :

A1 = 10 -10/2^1 = 10-5

A1 = 5

A2 = 10 -10/2^2 = 10-10/4 = 10-5/2 = (20-5)/2

A2 = 15/2

A3 = 10 -10/2^3 = 10-10/8 =10- 5/4 = (40-5)/4

A3 = 35/4

A4 = 10 -10/2^4 = 10-10/16 = 10-5/8 = (80-5)/8

A4 = 75/8

6 0

2 years ago

I need help on this math

goblinko [34]

I believe it is g(5) = 27 because if you look at the table in the x column when x=5, g(x)=27

3 0

3 years ago

A machine that fills beverage cans is supposed to put 12 ounces of beverage in each can. Following are the amounts measured in a

dlinn [17]

Answer:

The mean volume is 12 ounces.

Step-by-step explanation:

In this case we need to test whether the mean volume the machine fills the beverage cans with is 12 ounces or not.

The hypothesis for this test can be defined as:

H₀: The mean volume is 12 ounces. i.e. μ = 12 ounces.

Hₐ: The mean volume is not 12 ounces. i.e. μ ≠ 12 ounces.

As the population standard deviation is not known, we will use a t-test for single mean.

The sample mean and sample standard deviation is:

$\bar x=\frac{1}{n}\sum X\\=\frac{1}{8}\times [11.98+12.08+11.87+11.95+11.89+12.01+11.97+12.10]\\=11.98$

$\text{s} = \sqrt{\dfrac{\sum_{i=1}^{n}(x_i - \overline{x})^{2}}{n - 1}}\\=\sqrt{\dfrac{\sum_{i=1}^{n}(11.98 - 11.98)^{2}}{8 - 1}}+\sqrt{\dfrac{\sum_{i=1}^{n}(12.08- 11.98)^{2}}{8 - 1}}+...+\sqrt{\dfrac{\sum_{i=1}^{n}(12.10- 11.98)^{2}}{8 - 1}}\\=0.0815$ Compute the test statistic as follows: