All categories

3 years ago

Write 26 as repeated multiplication?

Mathematics

1 answer:

Harlamova29_29 [7]3 years ago

5 0

Answer:

This is your answer

Table of 26 gives the repeated addition of number 26, repeated for a certain number of times. For example, 4 baskets of 26 apples each, gives the sum of 26 + 26 + 26 + 26 = 104.

You might be interested in

At the dog park, 2 dogs are black with spots, 5 dogs are white with spots, 1 dog is solid white, and 4 dogs are solid black. Wha

Nikolay [14]

Its 11/12 , you need to divide the "wanted results" (every dog which is either black, with spots or both ) by every every possible result ( all dogs ). so 11/12=0.92

5 0

3 years ago

Read 2 more answers

Nemo's aquarium is filled with 2,4002{,}400 2,400 2, comma, 400 cubic centimeters of water. The base of the aquarium is 20 cm 20

kolezko [41]

The water in the aquarium is a rectangular prism with dimensions 20, 12 and h (where the height is unknown).

We know that the volume is the product of the dimensions, and that it is 2400, so we have

$V=2400=20\cdot 12 \cdot h \iff 240h=2400 \iff h=10$

6 0

3 years ago

What is the quotient of 8/9 and 1 1/3

Art [367]

Hello,

To divide by a fraction, multiply by its reciprocal.

Solution:

8/0 x 3/11

The greatest common factor is 3; cancel 3 out.

The problem will simplify to 8/33.

Faith xoxo

5 0

3 years ago

Mr. Snow bought 90 grams of Christmas candy for each of his 14 grandchildren. How many total kilograms of candy did he buy?

nignag [31]

Answer:

1.26 kg.

Step-by-step explanation:

We have been given that Mr. Snow bought 90 grams of Christmas candy for each of his 14 grandchildren. We are asked to find the amount of candy bought by Mr. Snow in kilograms.

First of all, we will find the amount of candy in grams by multiplying 90 grams by 14 as:

$\text{Amount of candy bought in grams}=14\times 90$

$\text{Amount of candy bought in grams}=1260$

We know that 1 kilogram is equal to 1000 grams. To convert 1260 grams into kg, we will divide 1260 by 1000 as:

$\text{Amount of candy bought in kilograms}=\frac{1260}{1000}$

$\text{Amount of candy bought in kilograms}=1.26$

Therefore, Mr. Snow bought 1.26 kilograms of candy.

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

Write 26 as repeated multiplication?​

Write 26 as repeated multiplication?