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ipn [44]
3 years ago
10

The diagram below.....​

Mathematics
1 answer:
vitfil [10]3 years ago
5 0

Answer:

what are the rest of the answer options I need to know that before I answer it

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HELP ME PLEASE!!! Asap <br><br> Solve for the proportion 42/21 = 6/d
MrMuchimi
D = 6/42121 or d = 0.000142447
7 0
3 years ago
Mary has 5 friends who are sharing a bag of chips with her. If there are 10 ounces in the bag and the girls share equally, how m
dybincka [34]

Answer:

1  1/4

Step-by-step explanation:

10/6=5/4

1   1/4

You divide the number of ounces by the number of people eating and the reduced answer is your final answer

8 0
3 years ago
Pls help me i really need it
babymother [125]

Answer:

The answer is below

Step-by-step explanation:

Let x represent the number of slides ridden and y represent the total cost. Hence:

1) For option 1, the table is shown. It passes through (2,10) and (6, 20). The equation is:

y-y_1=\frac{y_2-y_1}{x_2-x_1} (x-x_1)\\\\y-10=\frac{20-10}{6-2} (x-2)\\\\y=2.5x+5

2) For option 2, the equation is given by y = 5x + 7.5

3) For option 3, the graph passes through (30, 100) and (10, 40). Hence the equation is:

y-y_1=\frac{y_2-y_1}{x_2-x_1} (x-x_1)\\\\y-40=\frac{100-40}{30-10} (x-10)\\\\y=3x+10

Option 2 has the highest rate

7 0
3 years ago
If f(x) = -3 and g(x) = 3x2 + x - 6, find (f+ g)(x).
barxatty [35]

Answer:

3x^2+x-9

Step-by-step explanation:

f+g

means you are going to add whatever f equals to what g equals

so you have

(-3)+(3x^2+x-6)

Combine like terms

3x^2+x+(-3+-6)

3x^2+x+-9

or

3x^2+x-9

7 0
3 years ago
The mean and the standard deviation of a sampled population​ are, respectively, 113.9 and 32.1. Find the mean and standard devia
anzhelika [568]

Answer: The mean and standard deviation of the sampling distribution of the sample mean for samples of size n=64 is

113.9\text{ and }4.0 respectively.

Step-by-step explanation:

Given : The mean of sampled population : \mu = 113.9

The standard deviation of sampled population : \sigma = 32.1

We know that the mean and standard deviation of the sampling distribution of the sample mean for samples of size n is given by :_

\mu_s=\mu\\\\\sigma_s=\dfrac{\sigma}{\sqrt{n}}

Now, the mean and standard deviation of the sampling distribution of the sample mean for samples of size n=64 will be :-

\mu_s=113.9\\\\\sigma_s=\dfrac{32.1}{\sqrt{64}}=4.0125\approx4.0

7 0
3 years ago
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