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Which digit is in the ten thousandths place in 4,6037?

Iteru [2.4K]

Answer:

7

Step-by-step explanation:

4- Ones

6- tenths

0-hundredths

3-thousandths

7-ten-thousandths

Hope this helps! :)

6 0

3 years ago

Identify the missing base of the trapezoid, given that A=36 cm∧2. HELP PLEASE!!

den301095 [7]

Answer:

6 cm

Step-by-step explanation:

A = 1/2(b + 12)(4)

36 = 1/2(b + 12)(4)

72 = (b + 12)(4)

18 = b + 12

18 - 12 = b

6 = b

or

b = 6 cm

3 0

3 years ago

The manager of a grocery store selected 100 customers at random to ask what the maximum price is

Brilliant_brown [7]

What is the rest of the question?

8 0

3 years ago

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

1 + tanx / 1 + cotx =2

Lera25 [3.4K]

Answer:

x = tan^(-1)((i sqrt(3))/2 + 1/2) + π n_1 for n_1 element Z

or x = tan^(-1)(-(i sqrt(3))/2 + 1/2) + π n_2 for n_2 element Z

Step-by-step explanation:

Solve for x:

1 + cot(x) + tan(x) = 2

Multiply both sides of 1 + cot(x) + tan(x) = 2 by tan(x):

1 + tan(x) + tan^2(x) = 2 tan(x)

Subtract 2 tan(x) from both sides:

1 - tan(x) + tan^2(x) = 0

Subtract 1 from both sides:

tan^2(x) - tan(x) = -1

Add 1/4 to both sides:

1/4 - tan(x) + tan^2(x) = -3/4

Write the left hand side as a square:

(tan(x) - 1/2)^2 = -3/4

Take the square root of both sides:

tan(x) - 1/2 = (i sqrt(3))/2 or tan(x) - 1/2 = -(i sqrt(3))/2

Add 1/2 to both sides:

tan(x) = 1/2 + (i sqrt(3))/2 or tan(x) - 1/2 = -(i sqrt(3))/2

Take the inverse tangent of both sides:

x = tan^(-1)((i sqrt(3))/2 + 1/2) + π n_1 for n_1 element Z

or tan(x) - 1/2 = -(i sqrt(3))/2

Add 1/2 to both sides:

x = tan^(-1)((i sqrt(3))/2 + 1/2) + π n_1 for n_1 element Z

or tan(x) = 1/2 - (i sqrt(3))/2

Take the inverse tangent of both sides:

Answer: x = tan^(-1)((i sqrt(3))/2 + 1/2) + π n_1 for n_1 element Z

or x = tan^(-1)(-(i sqrt(3))/2 + 1/2) + π n_2 for n_2 element Z

4 0

3 years ago