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3 years ago

Choose all the fractions that are equivalent to 4/8 .

Mathematics

2 answers:

olga2289 [7]3 years ago

8 0

Answer:

can u show the options

Step-by-step explanation:

Rasek [7]3 years ago

8 0

Answer: 1/4 and 2/4

Step-by-step explanation:

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A frog 3” long can jump a distance of 18”. If a person 5 feet tall had the ability to jump like a frog, how far could he jump?

steposvetlana [31]

Answer:

130 inches

Step-by-step explanation:

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3 years ago

Write the equation in function form. Show your steps. 2x-3y=0

skad [1K]

Answer: Heyaa!

Your Answer is... \frac{3y}{2}

Step-by-step explanation:

Substitute the value of the variable into the equation and simplify.

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2 years ago

563 pints equals how many cups Answer:

Firdavs [7]

Answer:

1126 cups

Step-by-step explanation:

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3 years ago

Read 2 more answers

In the graph shown, suppose that f(x) = 3.5x and h(x) = 1.5x. Choose the statement that COULD be true.

faust18 [17]

Answer: D

Step-by-step explanation:

The given functions are

Now these are exponential curves and the bases for the functions are 3.5 & 1.5

Also the graph of g(x) is between f(x) & h(x)

Hence the value of base called the scale factor must be between 3.5 & 1.5.

4 & 5 are more than 3.5

0.9 is smaller than 1.5

But π = 3.14 lies between 3.5 & 1.5.

Hence the only option which can represent the graph of g(x) is

Option D) is the right answer

3 0

3 years ago

A teacher was interested in knowing the amount of physical activity that his students were engaged in daily. He randomly sampled

klasskru [66]

Answer:

The standard error of the mean is 4.5.

Step-by-step explanation:

As we don't know the standard deviation of the population, we can estimate the standard error of the mean from the standard deviation of the sample as:

$\sigma_{\bar{x}}\approx\frac{s}{\sqrt{n}}$

The sample is [30mins, 40 mins, 60 mins, 80 mins, 20 mins, 85 mins]. The size of the sample is n=6.

The mean of the sample is:

$\bar{x}=\frac{1}n} \sum x_i =\frac{30+40+60+80+20+85}{6}=52.5$

The standard deviation of the sample is calculated as:

$s=\sqrt{\frac{1}{n-1}\sum (x_i-\bar x)^2} \\\\ s=\sqrt{\frac{1}{5}\cdot ((30-52.5)^2+(40-52.5)^2+(60-52.5)^2+(80-52.5)^2+(20-52.5)^2+(85-52.5)^2}\\\\s=\sqrt{\frac{1}{5} *3587.5}=\sqrt{717.5}=26.8$

Then, we can calculate the standard error of the mean as:

$\sigma_{\bar{x}}\approx\frac{s}{\sqrt{n}}=\frac{26.8}{6}= 4.5$

6 0

3 years ago