All categories

2 years ago

Write an equivalent multiplication expression for 1/3 divided by 3

Mathematics

1 answer:

SpyIntel [72]2 years ago

3 0

Answer:

1/3 times 1/3

Step-by-step explanation:

the answer is 1/3 * 1/3 because when you divide with fraction you make the 3 or 3/1 turned around which is 1/3, then you multiply 1/3 * 1/3

You might be interested in

A rectangle has an area of 20 square units what can be said if it's height and width

Butoxors [25]

The height could be 2 and the width 10 or the height could be 10 and the width could be 2.

4 0

3 years ago

Read 2 more answers

In one day, a book store earned $253 in sales for 4 copies of a new cookbook and 5 copies of a new science fiction novel. On the

Aloiza [94]

Answer: cookbook $40 and science $15

Step-by-step explanation:

5 0

2 years ago

Read 2 more answers

Can you workout this problem

Vinil7 [7]

Hi,

Work:

Equation;

$35 - 3 \times {10}^{2} \div 20$
1. Write division as fraction.

$35 - \frac{3 \times {10}^{2} }{20}$

2. Evaluate the power of 10².

$35 - \frac{3 \times 100}{20}$

3. Reduce the fraction with 20.

$35 - \frac{3 \times 100}{20} = 35 - 3 \times 5$

4. Multiply.

$35 - 15$

5. Finally subtract and RESULT.

$35 - 15 = 20$

Hope this helps.
r3t40

7 0

3 years ago

A random sample of soil specimens was obtained, and the amount of organic matter (%) in the soil was determined for each specime

Shalnov [3]

Answer:

We conclude that the true average percentage of organic matter in such soil is different from 3%.

Step-by-step explanation:

We are given that the values of the sample mean and sample standard deviation are 2.481 and 1.616, respectively.

Suppose we know the population distribution is normal, we have to test the hypothesis that does this data suggest that the true average percentage of organic matter in such soil is something other than 3%.

Let $\mu$ = true average percentage of organic matter in such soil

SO, Null Hypothesis, $H_0$ : $\mu$ = 3% {means that the true average percentage of organic matter in such soil is equal to 3%}

Alternate Hypothesis, $H_A$ : $\mu \neq$ 3% {means that the true average percentage of organic matter in such soil is different than 3%}

The test statistics that will be used here is One-sample t test statistics because we don't know about the population standard deviation;

T.S. = $\frac{\bar X -\mu}{{\frac{s}{\sqrt{n} } } }$ ~ $t_n_-_1$

where, $\bar X$ = sample mean amount of organic matter = 2.481%

s = sample standard deviation = 1.616%

n = sample of soil specimens = 30

So, test statistics = $\frac{0.02481-0.03}{{\frac{0.01616}{\sqrt{30} } } }$ ~ $t_2_9$

= -1.759

Now, P-value of the test statistics is given by;

P-value = P( $t_2_9$ > -1.759) = 0.046 or 4.6%

If the P-value of test statistics is more than the level of significance, then we will not reject our null hypothesis as it will not fall in the rejection region.

If the P-value of test statistics is less than the level of significance, then we will reject our null hypothesis as it will fall in the rejection region.

Now, here the P-value is 0.046 which is clearly smaller than the level of significance of 0.05 (for two-tailed test), so we will reject our null hypothesis as it will fall in the rejection region.

Therefore, we conclude that the true average percentage of organic matter in such soil is different from 3%.

3 0

3 years ago

How do you solve, 2=2y-x and 23=5y- 4x , using the substitution method

kicyunya [14]

2= 2y - x
23 = 5y - 4x

isolate the x in the first equation

x = 2y - 2

now use this to substitute x in the second equation

23 = 5y - 4 (2y - 2)
23 = 5y - 8y + 6
23 = -3y + 6
-3y = 17
y = -5 2/3 (5.6 repeating )

now use that to find x in either equation.

2 = 2 (-5.66666) - x
2= -11.3 repeating - x
13.3 repeating = -x
x = -13.3 repeating

optional ** use the second equation to check

5 0

3 years ago

Write an equivalent multiplication expression for 1/3 divided by 3​

Write an equivalent multiplication expression for 1/3 divided by 3