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Mashutka [201]
2 years ago
8

What is 15267 times 16459 then divide by 7? ​

Mathematics
1 answer:
lesya [120]2 years ago
7 0

Answer:

33297327

Step-by-step explanation:

(15267 * 15267) ÷ 7

= 15267 * 15267

= 233081289 ÷ 7

= 33297327

Hope this helps! :)

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The total number of pears and oranges in a basket is between 30 and 40 . There are 25% more apple than orange how many apples ar
Verdich [7]

The number of apples there are in the basket as described in the task content is; 20 apples.

<h3>What is the number of apples there are in the basket?</h3>

Inference drawn from the task content show that;

  • The ratio of apples to oranges in the basket is; 5:4. making the total partitions be 9.

On this note, we must identify a number divisible by 9 between 30 and 40.

The number is 36 and once divided by 9 is 4 units per partition.

On this note, the total number of apples in the basket is; 4 × 5 = 20 apples.

Read more on ratios;

brainly.com/question/13513438

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7 0
2 years ago
Mr.Brookins asked his students to plot an interger on a number line. Which two students could have represented the same number?
VikaD [51]

Answer:

I need a picture

Step-by-step explanation:

8 0
2 years ago
What's 6x to the tenth power divided by12x to the second power
ehidna [41]

Answer:

\frac{1}{2} x⁸

Step-by-step explanation:

Given expression:

       \frac{6x^{10} }{12x^{2} }  

= \frac{6}{12}  x \frac{x^{10} }{x^{2} }  

   As a rule of exponents;

         \frac{a^{n} }{a^{m} }  = a^{n-m}

So;

            =  \frac{1}{2} x x^{10-2}  

            = \frac{1}{2} x⁸

4 0
3 years ago
Stella rewrites −212+3.7 as 3.7−212. Which property did she use? A. Additive Identity Property B. Additive Inverse Property C. C
lyudmila [28]

Answer:

Option C.

Step-by-step explanation:

It is given that Stella rewrites −212+3.7 as 3.7−212.

We need to find which property did she use.

According to commutative property of addition,

A+B=B+A

Where, A and B are any real values.

In the given expression −212+3.7, A=-212 and B=3.7.

Using commutative property of addition, we get

(-212)+3.7=3.7+(-212)

(-212)+3.7=3.7-212

So, she used commutative property.

Therefore, the correct option is C.

5 0
2 years ago
Archaeologists can determine the diets of ancient civilizations by measuring the ratio of carbon-13 to carbon-12 in bones found
Vsevolod [243]

Answer:

Step-by-step explanation:

The null and the alternative hypothesis are:

H₀=μ₀=μ₁=μ₂=μ₃=μ₄

H₁=two or more μ are different X

Let X_{ij} denote the jth observation of the ith sample. The mean and the variance of the ith sample is given by

Xbar_{i} =\frac{1}{J_{i}} summation(X_{ij}) where J=1 to J_{i} and  J_{i}  is ith sample size

s_{i} ^{2} =\frac{1}{J_{i}-1 }summation (X_{ij} -   Xbar_{i})^{2}

The sample sizes are J₁ =12   J₂=10   J₃=18   J₄=9

the total number in all samples combined is  49

finding Xbar₁ and s₁

Xbar₁= 1÷12(17.2+....+13.4)

Xbar₁= 17.0500

s₁²= 1÷(12-1) [(17.0500-17.2)+...+(17.0500-13.4)]

s₁²=5.1336

Similarly find the means and variances of other samples

\left[\begin{array}{ccc}i&Mean&variance\\1&17.0500&5.1336\\2&15.4500&13.2317\\3&16.4972&6.9460\\4&15.4444&7.4428\end{array}\right]

the sample grand mean denoted by Xbar is the average of all sampled items taken together:

Xbar=\frac{1}{49} (17.2+.....+16.7)

Xbar=16.2255

Compute the treatment sum of squares SSTr, the sum of squares SSE and the total sum of squares SST

SSTr = ∑ J_{i} (Xbar_{i} -Xbar)^{2}  from i=1 to 49

SSTr= 20.9910

SSE= \sum\limits^I_{i=1}\sum\limits^{J_i}_{j=1}(Xbar_{ij}-Xbar_i)^{2}

SSE=\sum\limits^I_{i=1}(J_i-1)s_i^{2}

SSE=(12-1)(5.1336)+...(9-1)(7.4428)

SSE=353.1796

SST=SSTr+SSE

SST=374.1706

Find the treatment mean square MSTr and the error mean square MSE:

MSTr= SSTr/(I-1)

MSTr=6.9970

MSE=SSE/(N-I)

MSE=7.8484

F=\frac{MSTr}{MSE}

F=0.89

The degrees of freedom are I-1=3 for the numerator and N-I=45 for the denominator. Under H₀, F has an F_{3,45} distribution. To find the P value we consult the F table.

P>0.100

The complete ANOVA table is below

\left[\begin{array}{cccccc}source&DF&SS&MS&F&P\\Agegroup&3&20.9910&6.9970&0.89&0.453\\Error&45&353.1796&7.8484\\Total&48&374.1706\end{array}\right]

(b) The P-value is large. A follower of 5% rule would not reject H₀. Therefore, we cannot conclude the concentration ratios differ among the age groups

4 0
3 years ago
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