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Alexus [3.1K]
3 years ago
15

I did the other problems but I dont know these can someone please help me?

Mathematics
2 answers:
suter [353]3 years ago
4 0

Answer:

░░░░░▐▀█▀▌░░░░▀█▄░░░

░░░░░▐█▄█▌░░░░░░▀█▄░░

░░░░░░▀▄▀░░░▄▄▄▄▄▀▀░░

░░░░▄▄▄██▀▀▀▀░░░░░░░

░░░█▀▄▄▄█░▀▀░░

░░░▌░▄▄▄▐▌▀▀▀░░ This is Bob

▄░▐░░░▄▄░█░▀▀ ░░

▀█▌░░░▄░▀█▀░▀ ░░ Copy And Paste Him onto all of ur brainly answers

░░░░░░░▄▄▐▌▄▄░░░ So, He Can Take

░░░░░░░▀███▀█░▄░░ Over brainly

░░░░░░▐▌▀▄▀▄▀▐▄░░

░░░░░░▐▀░░░░░░▐▌░░

░░░░░░█░░░░░░░░█░

do it or he will hunt you down and kill u (lets destroy the moderators!!!!!!!!)

we are slowing them down already! good work soilders!

Step-by-step explanation:

Marta_Voda [28]3 years ago
3 0

Answer:

#1 Equation: 36 + 4 = 8x  (40 = 8x) Solution: $5

#2 Equation: 36 - 8 Solution: 28

Step-by-step explanation:

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Two alloys A and B are being used to manufacture a certain steel product. An experiment needs to be designed to compare the two
sveta [45]

Answer:

Step-by-step explanation:

From the given information, we can compute the table showing the summarized statistics of the two alloys A & B:

                                            Alloy A        Alloy B

Sample mean                    \bar {x} _A = 49.5      \bar {x} _B = 45.5

Equal standard deviation    \sigma_A = 5         \sigma_B= 5

Sample size                          n_A = 30        n_{B}= 30

Mean of the sampling distribution is :

\mu_{\bar{X_1}-\bar{X_2}}= 49.5-45.5  \\ \\ = 4.0

Standard deviation of sampling distribution:

\sigma_{\bar{x_1}-\bar{x_2} }= \sqrt{\sigma^2_{\bar{x_1}-\bar{x_2} }} \\ \\ = \sqrt{\dfrac{\sigma_1^2}{n_1} + \dfrac{\sigma^2_2}{n_2} } \\ \\ =\sqrt{\dfrac{25}{30}+\dfrac{25}{30}} \\\\=\sqrt{1.667}  \\ \\ =1.2909

Hypothesis testing.

Null hypothesis:  H_o: \mu_A -\mu_B = 0

Alternative hypothesis:  H_A:\mu_A -\mu_B > 4

The required probability is:

P(\overline X_A - \overline X_B>4|\mu_A - \mu_B) = P\Big (\dfrac{(\overline X_A - \overline X_B)-\mu_{X_A-X_B}}{\sigma_{\overline x_A -\overline x_B}} > \dfrac{4 - \mu_{X_A-\overline X_B}}{\sigma _{\overline x_A - \overline X_B}}   \Big) \\ \\ = P \Big( z > \dfrac{4-0}{1.2909}\Big) \\ \\ = P(z \ge 3.10)\\ \\ = 1 - P(z < 3.10) \\ \\ \text{Using EXCEL Function:} \\ \\  = 1 - [NORMDIST(3.10)]  \\ \\ = 1- 0.999032 \\ \\ 0.000968 \\ \\ \simeq  0.0010

This implies that a minimal chance of probability shows that the difference of 4 is not likely, provided that the two population means are the same.

b)

Since the P-value is very small which is lower than any level of significance.

Then, we reject H_o and conclude that there is enough evidence to fully support alloy A.

3 0
3 years ago
Determine the interquartile range for the data.<br> 16, 19, 25, 20,<br> 22, 21, 17, 20
Roman55 [17]

Answer:

the answer is 15 subtract the highest number from the lowest

5 0
3 years ago
.......Help Please.........
Sveta_85 [38]

Answer:

57% to the nearest whole number

Step-by-step explanation:

That is (80/139.3) * 100

= 0.5743  * 100

= 57.43%.

8 0
3 years ago
A cafe owner wants to know how many cold drinks she is likely to sell next week
Simora [160]

she would first calculate how many drinks she sell the other weeks  and  know the rate at which each customer buy the drinks.

3 0
4 years ago
Plz Help ASAP!!!!!!!!
Setler [38]

Answer:

D

Step-by-step explanation:

the circle is filled in

7 0
4 years ago
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