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Nitella [24]
3 years ago
5

Help me plz 50 points and ill give brainliest if you answer all​

Mathematics
2 answers:
Masteriza [31]3 years ago
7 0

Answer:

11) 2x=-4

Divide both sides by 2:

\frac{2x}{2}=\frac{-4}{2}

x=-2

12) 2^3\cdot \:2^1=2^{3+1}=2^4

(2^4=16) if you wanted the exact value.

13) -3+10=x

Switch sides:

x=-3+10

Rearrange:

x=10-3

x=7

14) 10^2-64

10*10-64

100-64

36

15) \sqrt{100}-5^2

10-5^2

10-5*5

10-25

-15

16) \frac{1}{4}x=-10

Multiply both sides by 4:

4\cdot \frac{1}{4}x=4\left(-10\right)

x=-40

17) \left(2^3\right)^2=2^{3*2}=2^6

(2^6=64) if you wanted the exact value.

18) \left(2^3\right)\left(2^2\right)=2^{3+2}=2^5

(2^5=32) if you wanted the exact value.

Bonus Question:

Terrorist attacks took place against the United States in September 11 2001.

This is named "September 11 attacks" and you can search more about it online.

noname [10]3 years ago
6 0

Answer:

10. x = -2

12. 16

13. x = 7

14. 36

15. -15

16.  x = -40

17. 64

18. 32

Friday 9/11 The plane hijackings that struck the World Trade Center, the Pentagon, and a Pennsylvania field killed 2,977 people.

Step-by-step explanation:

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Please help. Will give brainliest and would really appreciate it.
kondor19780726 [428]

Answer:

x≠3

Step-by-step explanation:

3 0
3 years ago
Land's Bend sells a wide variety of outdoor equipment and clothing. The company sells both through mail order and via the intern
Naya [18.7K]

Answer:

80% confidence interval for the true mean difference between the mean amount of mail-order purchases and the mean amount of internet purchases (\mu_1-\mu_2) is [-9.132 , 23.332].

Step-by-step explanation:

We are given that a random sample of 7 sales receipts for mail-order sales results in a mean sale amount of $81.70 with a standard deviation of $18.75.

A random sample of 11 sales receipts for internet sales results in a mean sale amount of $74.60 with a standard deviation of $28.25.

Firstly, the Pivotal quantity for 80% confidence interval for the difference between population means is given by;

                            P.Q. =  \frac{(\bar X_1-\bar X_2)-(\mu_1-\mu_2)}{s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2} } }  ~ t__n__1-_n__2-2

where, \bar X_1 = sample mean sales receipts for mail-order sales = $81.70

\bar X_2 = sample mean sales receipts for internet sales = $74.60

s_1 = sample standard deviation for mail-order sales = $18.75

s_2 = sample standard deviation for internet sales = $28.25

n_1 = size of sales receipts for mail-order sales = 7

n_2 = size of sales receipts for internet sales = 11

Also, s_p=\sqrt{\frac{(n_1-1)s_1^{2} +(n_2-1)s_2^{2} }{n_1+n_2-2} } = \sqrt{\frac{(7-1)\times 18.75^{2} +(11-1)\times 28.25^{2} }{7+11-2} } = 25.11

<em>Here for constructing 80% confidence interval we have used Two-sample t test statistics as we don't know about population standard deviations.</em>

<em />

So, 80% confidence interval for the difference between population means, (\mu_1-\mu_2) is ;

P(-1.337 < t_1_6 < 1.337) = 0.80  {As the critical value of t at 16 degree

                                         of freedom are -1.337 & 1.337 with P = 10%}  

P(-1.337 < \frac{(\bar X_1-\bar X_2)-(\mu_1-\mu_2)}{s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2} } } < 1.337) = 0.80

P( -1.337 \times {s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2} } } < {(\bar X_1-\bar X_2)-(\mu_1-\mu_2)} < 1.337 \times {s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2} } } ) = 0.80

P( (\bar X_1-\bar X_2)-1.337 \times {s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2} } } < (\mu_1-\mu_2) < (\bar X_1-\bar X_2)+1.337 \times {s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2} } } ) = 0.80

<u>80% confidence interval for</u> (\mu_1-\mu_2) =

[ (\bar X_1-\bar X_2)-1.337 \times {s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2} } } , (\bar X_1-\bar X_2)+1.337 \times {s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2} } } ]

= [ (81.70-74.60)-1.337 \times {25.11 \times \sqrt{\frac{1}{7} +\frac{1}{11} } } , (81.70-74.60)+1.337 \times {25.11 \times \sqrt{\frac{1}{7} +\frac{1}{11} } } ]

= [-9.132 , 23.332]

Therefore, 80% confidence interval for the true mean difference between the mean amount of mail-order purchases and the mean amount of internet purchases (\mu_1-\mu_2) is [-9.132 , 23.332].

4 0
3 years ago
Does anyone know this? No links
Leni [432]

Answer:

The answer will be x = 3, you are correct!

Step-by-step explanation:

Hope this helps!! :)))

5 0
3 years ago
Read 2 more answers
7th grade math help me plzzz
matrenka [14]

Answer:

a. -14

b. -150

c. -51

d. -2

e. 3

f. 5

8 0
3 years ago
participar de uma maratona um atleta inicia um treinamento mensal, em que corre todo dia e sempre 2 minutos a mais do que correu
AleksandrR [38]

Answer:

59 minutes

Step-by-step explanation:

In the table of values ​​it is observed that for each day of training 2 minutes are increasing and that day 1 begins with 5 minutes of training , so

y(x)=2\frac{min}{d}(x)+3min, where x= evaluation day

we have to for the day 28

y(28)=2\frac{min}{d}(28d)+3min=59min

7 0
3 years ago
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