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GaryK [48]
3 years ago
14

Whst is 15 divided by 6 2/3 equals

Mathematics
2 answers:
PolarNik [594]3 years ago
5 0
First you make 15 a fraction by putting it over 1.
Then you make 6 2/3 a mixed fraction.
To do this, you multiply 6 with 3 and add 2 to it. Then you put the number over 3.
We get
6 X 3 = 18
18 +2 = 20
20/3

Now, we divide 15/1 with 20/3.
When we divide it, we actually cross multiply it. Basically, we flip the fraction and multiply it.

So we get 15/1 X 3/20 = \frac{15 X 3}{1 X 20}.

We get 45/20 which can be simplified to 9/5 which can be simplified to 2  1/4.

So the answer is 9/4 or 2  1/4.
Fed [463]3 years ago
5 0
If you would like to know what is 15 divided by 6 2/3, you can calculate this using the following steps:

6 2/3 = 20/3

15 / 20/3 = 15 * 3/20 = 45/20 = 9/4 = 2.25

Result: 15 divided by 6 2/3 equals to 9/4.
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Can someone plz help me with Algebra 2....... :) Will give Brainliest!!
Anarel [89]
Using logarithms property of log(x)+log(y)=log(xy)
so here, you can sum the equation to;
log((x+6)*(x-6))=2
so you can simply say that;
log_{8}((x+6)( x-6))=2
and by multiplying (x+6)*(x-6)
log_{8}(x^2-36)=2
and as you know also that;
a^{b}=c is same as log _{a}c=b
so you can simply state it as;
8^2=x^2-36
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x^2=100
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Hope this helps!
6 0
3 years ago
1/4 and 7/8 as an expression
Maslowich

Answer:

1/4: 1s/4

7/8: 7x/s

Step-by-step explanation:

3 0
3 years ago
Read 2 more answers
Help this math problem I can't solve it
Tcecarenko [31]
To answer this problem, please subtract 5 1/3 mile from 12 1/3 mile.

That'd be 

       12 1/3
   -     5 1/3
-----------------
         7 mi
5 0
3 years ago
A person drilled a hole in a die and filled it with a lead​ weight, then proceeded to roll it 200 times. Here are the observed f
Alona [7]

Solution :

The objective of the study is to test the claim that the loaded die behaves at a different way than a fair die.

Null hypothesis, $H_0:p_1=p_1=p_3=p_4=p_5=p_6=\frac{1}{6}$

That is the loaded die behaves as a fair die.

Alternative hypothesis, $H_a$ : loaded die behave differently than the fair die.

Number of attempts , n = 200

Expected frequency, $E_i=np_i$

                                        $=200 \times \frac{1}{6} = 33.333$

Test statistics, $x^2= \sum^6_{i=1} \frac{(O_i-E_i)^2}{E_i} $

                            $=\frac{(28-33.333)^2}{33.333}+\frac{(29-33.333)^2}{33.333}+\frac{(40-33.333)^2}{33.333}+\frac{(41-33.333)^2}{33.333}+\frac{(28-33.333)^2}{33.333}+$$\frac{(34-33.333)^2}{33.333}$

≈ 5.8

Degrees of freedom, df = n - 1

                                       = 6 - 1

                                       = 5

Level of significance, α = 0.10

At α = 0.10 with df = 5, the critical value from the chi square table

$x^2_{\alpha}= \text{chi inv}(0.10,5)$

     = 9.236

Thus the critical value is $x_{\alpha}^2=9.236$

$P \text{ value} = P[x^2_{df} \geq x^2]$

             $=P[x^2_5\geq 5.80]$

             = chi dist (5.80, 5)

             = 0.3262

Decision : The value of test statistics 5.80 is not greater than the critical value 9.236, thus fail to reject $H_o$ at 10% LOS.

Conclusion : There is no enough evidence to support the claim that the loaded die behave in a different than a fair die.

3 0
3 years ago
Find the value of z and x?
zloy xaker [14]

Answer:

For z I don't know what to do but thats for x, maybe someone else could find z for you :)

Step-by-step explanation:

5x - 89 = 61

      +89    +89

----------------------

5x = 50

----    -----

5         5

x = 10

6 0
3 years ago
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