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Anna11 [10]
3 years ago
15

HELP ME PLEASE WITH THESE MATH PROBLEMS AHHHHHHHHHHH

Mathematics
1 answer:
djyliett [7]3 years ago
4 0

Answer: You can simply use a calculator.

Step-by-step explanation:

You might be interested in
PLEASE HELP RIGHT AWAY!!!!!
Cloud [144]

Answer:

1) 46.6

Step-by-step explanation:

Given Data:

s1 of Triangle ABC= 4cm

s2 of Triangle ABC=6cm

s3 of Triangle ABC=8cm

As per the side-side-side SSS of a given triangle

the corresponding angles are found by the following law of cosine formula

cos A = (b2 + c2 − a2) / 2bc

Putting the values we get

cos A= (4^2+8^2-6^2)/2(4)(8)

cos A= (16+64-36)/64

A=cos^-0.6875

A=46.57!

4 0
3 years ago
Solve the equation for m. Show your work -40 - 6m + 1 = - 2
bekas [8.4K]

Answer:

\bold{m \ = \ \frac{-37}{6} \ Or  \ -6.16666666667}

Explanation:

-40 - 6m + 1 = - 2

  • Simplify Both Sides Of The Equation

−40 − 6m + 1 = −2

−40 + −6m + 1 = −2

(−6m) + (−40 + 1) = −2 (Combine Like Terms)

−6m + −39 = −2

= −6m − 39 = −2

  • Add 39 To Both Sides

−6m − 39 + 39 = −2 + 39

−6m = 37

  • Divide Both Sides By -6

-6m / -6 = 37 / -6

  • Simplify

\bold{m \ = \ \frac{-37}{6} \ Or  \ -6.16666666667}

- PNW

4 0
3 years ago
Evaluate the expression for the value 5 X (h + 3) for h= 7
horrorfan [7]

Answer:

35x + 15

Step-by-step explanation:

5x(7 + 3)

35x + 15

8 0
3 years ago
Read 2 more answers
A simulation was conducted using 10 fair six-sided dice, where the faces were numbered 1 through 6. respectively. All 10 dice we
kompoz [17]

Answer:

C) a sample distribution of a sample mean with n = 10  

\mu_{{\overline}{X}} = 3.5

and \sigma_{{\overline}{Y}} = 0.38

Step-by-step explanation:

Here, the random experiment is rolling 10, 6 faced (with faces numbered from 1 to 6) fair dice and recording the average of the numbers which comes up and the experiment is repeated 20 times.So, here sample size, n = 20 .

Let,

X_{ij} = The number which comes up  on the ith die on the jth trial.

∀ i = 1(1)10 and j = 1(1)20

Then,

E(X_{ij}) = \frac {1 + 2 + 3 + 4 + 5 + 6}{6}

                            = 3.5       ∀ i = 1(1)10 and j = 1(1)20

and,

E(X^{2}_{ij} = \frac {1^{2} + 2^{2} + 3^{2} + 4^{2} + 5^{2} + 6^{2}}{6}

                                = \frac {1 + 4 + 9 + 16 + 25 + 36}{6}

                                = \frac {91}{6}

                                \simeq 15.166667

so, Var(X_{ij} = (E(X^{2}_{ij} - {(E(X_{ij})}^{2})

                                    \simeq 15.166667 - 3.5^{2}

                                    = 2.91667

   and \sigma_{X_{ij}} = \sqrt {2.91667}[/tex                                            [tex]\simeq 1.7078261036

Now we get that,

 Y_{j} = \frac {\sum_{j = 1}^{20}X_{ij}}{20}

We get that Y_{j}'s are iid RV's ∀ j = 1(1)20

Let, {\overline}{Y} = \frac {\sum_{j = 1}^{20}Y_{j}}{20}

      So, we get that E({\overline}{Y}) = E(Y_{j})

                                                                 = E(X_{ij}  for any i = 1(1)10

                                                                 = 3.5

and,

       \sigma_{({\overline}{Y})} = \frac {\sigma_{Y_{j}}}{\sqrt {20}}                                             = \frac {\sigma_{X_{ij}}}{\sqrt {20}}                                             = \frac {1.7078261036}{\sqrt {20}}                                            [tex]\simeq 0.38

Hence, the option which best describes the distribution being simulated is given by,

C) a sample distribution of a sample mean with n = 10  

\mu_{{\overline}{X}} = 3.5

and \sigma_{{\overline}{Y}} = 0.38

                                   

6 0
3 years ago
First, rewrite<br> 7/9 and 16/21<br> so that they have a common denominator.
sattari [20]
49/63 and 48/63..........
5 0
3 years ago
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