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Vikentia [17]
3 years ago
5

10 3/4*5 1/4*9 1/2 as a mixed number

Mathematics
1 answer:
laiz [17]3 years ago
7 0

Answer:

43/4.    21/4.   19/2

Step-by-step explanation:

You multiply the decimator and add the nunimator

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Evaluate the expression when x=-5.<br> Huiles<br> x2 +8x+2
IgorC [24]

Answer:

52

Step-by-step explanation:

(2×5)+(8×5)+2

10+40+2

52

5 0
3 years ago
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Solve 227 students went on a fieldtrip. 9 buses were filled and 12 students rode in cars. which equation can be used to find p,
nasty-shy [4]
9p+12=227
9 bus time p person per bus equal all people riding buses

Is there a choice like this?
6 0
3 years ago
What is the binomial expansion of (2x – 3)^5?
Tanya [424]

Answer:

C

Step-by-step explanation:

(2x + 3)^5 = C(5,0)2x^5*3^0 +

C(5,1)2x^4*3^1 + C(5,2)2x^3*3^2 + C(5,3)2x^2*3^3 + C(5,4)2x^1*3^4 + C(5,5)2x^0*3^5

Recall that

C(n,r) = n! / (n-r)! r!

C(5,0) = 1

C(5,1) = 5

C(5,2) = 10

C(5,3) = 10

C(5,4) = 5

C(5,5) = 1

= 1(2x^5)1 + 5(2x^4)3 + 10(2x^3)3^2 + 10(2x^2)3^3 + 5(2x^1)3^4 + 1(2x^0)3^5

= 2x^5 + 15(2x^4) + 90(2x^3) + 270(2x^2) + 405(2x) +243

= 32x^5 + 15(16x^4) + 90(8x^3) + 270(4x^2) + 810x + 243

= 32x^5 + 240x^4 + 720x^3 + 1080x^2 + 810x + 243

7 0
3 years ago
Read 2 more answers
Jacob Lee is a frequent traveler between Los Angeles and San Diego. For the past month, he wrote down the flight times in minute
valkas [14]

Solution :

We know that

$H_0: \mu_1 = \mu_2=\mu_3$

$H_1 :$ At least one mean is different form the others (claim)

We need to find the critical values.

We know k = 3 , N = 35, α = 0.05

d.f.N = k - 1

       = 3 - 1 = 2

d.f.D = N - k

        = 35 - 3 = 32

SO the critical value is 3.295

The mean and the variance of each sample :

Goust                      Jet red                 Cloudtran

$\overline X_1 =50.5$           $\overline X_2 =50.07143$        $\overline X_3 =55.71429$

$s_1^2=19.96154$      $s_2^2=14.68681$         $s_3^2=36.57143$

The grand mean or the overall mean is(GM) :

$\overline X_{GM}=\frac{\sum \overline X}{N}$

         $=\frac{51+51+...+49+49}{35}$

        = 52.1714

The variance between the groups

$s_B^2=\frac{\sum n_i\left( \overline X_i - \overline X_{GM}\right)^2}{k-1}$

     $=\frac{\left[14(50.5-52.1714)^2+14(52.07143-52.1714)^2+7(55.71426-52.1714)^2\right]}{3-1}$

   $=\frac{127.1143}{2}$

   = 63.55714

The Variance within the groups

$s_W^2=\frac{\sum(n_i-1)s_i^2}{\sum(n_i-1)}$

    $=\frac{(14-1)19.96154+(14-1)14.68681+(7-1)36.57143}{(14-1)+(14-1)+(7-1)}$

   $=\frac{669.8571}{32}$

  = 20.93304

The F-test  statistics value is :

$F=\frac{s_B^2}{s_W^2}$

  $=\frac{63.55714}{20.93304}$

  = 3.036212

Now since the 3.036 < 3.295, we do not reject the null hypothesis.

So there is no sufficient evidence to support the claim that there is a difference among the means.

The ANOVA table is :

Source       Sum of squares    d.f    Mean square    F

Between    127.1143                 2      63.55714          3.036212

Within        669.8571             32      20.93304

Total           796.9714            34

3 0
3 years ago
Finish the following table for the given function with x as thr independent variable.
konstantin123 [22]
The answer is c.0.1,0.05.0.025.
We can find these by substituting the x values on the table given into the equation; h(x)=1/x.
5 0
3 years ago
Read 2 more answers
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