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son4ous [18]
4 years ago
15

Can someone help me I forgot how to do this

Mathematics
1 answer:
Yuliya22 [10]4 years ago
8 0
Perimeter = 7+5+7+5 = 24
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Help please! im confused
kati45 [8]

Answer:

1/2

Step-by-step explanation:

area of larger square: 12 x 12 = 144

area of smaller square: 6√2 x 6√2 = 72

S/B = 72/144 = 1/2

4 0
3 years ago
The sum of a number and 4 times its reciprocal is 13/3<br> Find both possible numbers.
Firdavs [7]

Answer:

3 ; 4/3

Step-by-step explanation:

x + 4(1/x) = 13/3

x + 4/x = 13/3

x^2 -13/3x + 4 = 0

3x^2 - 13x + 12 = 0

Δ = 169 - 144 = 25

x1 = (13 + 5)/6 = 3

x2 = (13-5)/6 = 4/3

4 0
3 years ago
Read 2 more answers
The sum of a number times 3 and 6 equals 4
victus00 [196]
Let a number = x
therefore the sum of a number times 3 = 3x
and 6 means 3x+6 equals 4
so the equation is 3x+6=4 
3 0
4 years ago
Chloe breaks into the pantry after Albert goes to work. She finds and opens a family-pack of mixed nuts. After surfing the web,
Firdavs [7]

Solution :

In this case we have to use $\text{chi square test}$ for the goodness of fit.

Null hypothesis is : $H_0 :$ Data follows the given distribution

Alternate hypothesis is : $H_a:$ Data do not follow the given distribution

The level of significance, $\alpha = 0.05$

The test statistics is given by :

Chi square = $\sum\frac{(O-E)^2}{E}$

Here O is the observed frequencies

         E is the expected frequencies

So we have, N = number of the categories = 4

df = degrees o freedom = N - 1

                                        = 4 - 1 = 3

Critical value  = $7.814727764$

Calculating the table for the test statistics are given as :

  Category      prop.     O        E       $\frac{(O-E)^2}{E}$

   A                   0.2     14       20         1.80

   C                   0.3     28       30         0.13

   M                  0.1        6        10          1.60

  P                    0.4      52       40         3.60

Total                  1        100      100        7.13

The test statics = chi square  = $\sum\frac{(O-E)^2}{E}$ = 7.13

$X^2$ statistics = 7.13

P-value = 0.067767248

P -value > $\alpha = 0.05$

Therefore, we do not reject the $\text{null hypothesis}$. There is sufficient evidence to conclude that the data follows the given distribution. So there is sufficient evidence to conclude the given claim is true.

4 0
3 years ago
Open the picture from the other question and do it here​
Alik [6]

k(x) - g(x) + f(x)

(k(x) + f(x))  - g(x)

((2x + 5) + (x - 1)) - ( {6x}^{2}  - 4x +  {3x}^{4}

(2x + 5 +x - 1)  - 6 {x}^{2}   + 4x  -  {3x}^{4}

2x + 5 +x - 1  - 6 {x}^{2}   + 4x  -  {3x}^{4}

( -  {3x}^{4} ) + ( -  {6x}^{2} ) + (2x + x + 4x) + (5 - 1)

-  {3x}^{4} -  {6x}^{2}  + 7x  + 4

4 0
3 years ago
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