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2 years ago

∠A and ∠C are right angles. m∠p = _???_ degrees.

Mathematics

2 answers:

iogann1982 [59]2 years ago

5 0

Answer

Find out the m∠p .

To prove

As in ΔDAB is a right triangle

Apply pythagorean theorem

Hypotenuse ² = Perpendicular ² + Base²

DB² = AB² + AD²

AB = 5 units

AD = 6 units

Put in the above formula

DB² = 5² + 6²

= 25 + 36

= 61

$DB = \sqrt{61}\ units$

= 7.8 units (approx)

Now in ΔDCB is a right triangle .

By using the trignometric identity .

$cosp = \frac{Base}{Hypotenuse}$

$cosp = \frac{DC}{DB}$

As DC = 4 units

DB = 7.8 units (approx)

Put all the values in the trignometric identity .

$cos p = \frac{4}{7.8}$

$\angle p = cos^{-1}(\frac{4}{7.8})$

∠p = 59.15 ° (approx)

goblinko [34]2 years ago

4 0

Hope this helped you.

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Which of the following functions is graphed below?

dalvyx [7]

Answer:

The function graphed is y = Ix - 5I - 4 ⇒ answer D

Step-by-step explanation:

* Lets explain how to solve this problem

- From the graph

# The graph intersects the x-axis at x = 1 and x = 9

∴ The x-intercepts are 1 and 9

- We can find x-intercept by equate y by 0

* Lets equate the answers by zero to find the x-intercept

# y = Ix + 5I - 4

∵ Ix + 5I - 4 = 0

- Add 4 to both sides

∴ Ix + 5I = 4

- Remember in Ia + bI = c, then we have two answers :

a + b = c <em>OR </em> a + b = -c

∵ Ix + 5I = 4

∴ x + 5 = 4 ⇒ subtract 5 from both sides

∴ x = -1

- OR

∴ x + 5 = -4 ⇒ subtract 5 from both sides

∴ x = -9

∴ The x-intercepts are -1 and -9 not the same with figure

# y = Ix - 5I + 4

∵ Ix - 5I + 4 = 0

- Subtract 4 from both sides

∴ Ix - 5I = -4

- Remember in Ia + bI = c , c can't be negative because the absolute

value is always positive

∴ We can't solve this equation

# y = Ix + 5I + 4

∵ Ix + 5I + 4 = 0

- Subtract 4 from both sides

∴ Ix + 5I = -4

- Remember in Ia + bI = c , c can't be negative because the absolute

value is always positive

∴ We can't solve this equation

# y = Ix - 5I - 4

∵ Ix - 5I - 4 = 0

- Add 4 to both sides

∴ Ix - 5I = 4

- Remember in Ia + bI = c, then we have two answers :

a + b = c <em>OR </em> a + b = -c

∵ Ix - 5I = 4

∴ x - 5 = 4 ⇒ add 5 to both sides

∴ x = 9

- OR

∴ x - 5 = -4 ⇒ add 5 to both sides

∴ x = 1

∴ The x-intercepts are 1 and 9 the same with figure

* The function graphed is y = Ix - 5I - 4

5 0

2 years ago

Two machines are used for filling glass bottles with a soft-drink beverage. The filling process have known standard deviations s

stellarik [79]

Answer:

a. We reject the null hypothesis at the significance level of 0.05

b. The p-value is zero for practical applications

c. (-0.0225, -0.0375)

Step-by-step explanation:

Let the bottles from machine 1 be the first population and the bottles from machine 2 be the second population.

Then we have $n_{1} = 25$ , $\bar{x}_{1} = 2.04$ , $\sigma_{1} = 0.010$ and $n_{2} = 20$ , $\bar{x}_{2} = 2.07$ , $\sigma_{2} = 0.015$ . The pooled estimate is given by

$\sigma_{p}^{2} = \frac{(n_{1}-1)\sigma_{1}^{2}+(n_{2}-1)\sigma_{2}^{2}}{n_{1}+n_{2}-2} = \frac{(25-1)(0.010)^{2}+(20-1)(0.015)^{2}}{25+20-2} = 0.0001552$

a. We want to test $H_{0}: \mu_{1}-\mu_{2} = 0$ vs $H_{1}: \mu_{1}-\mu_{2} \neq 0$ (two-tailed alternative).

The test statistic is $T = \frac{\bar{x}_{1} - \bar{x}_{2}-0}{S_{p}\sqrt{1/n_{1}+1/n_{2}}}$ and the observed value is $t_{0} = \frac{2.04 - 2.07}{(0.01246)(0.3)} = -8.0257$ . T has a Student's t distribution with 20 + 25 - 2 = 43 df.

The rejection region is given by RR = {t | t < -2.0167 or t > 2.0167} where -2.0167 and 2.0167 are the 2.5th and 97.5th quantiles of the Student's t distribution with 43 df respectively. Because the observed value $t_{0}$ falls inside RR, we reject the null hypothesis at the significance level of 0.05

b. The p-value for this test is given by $2P(T$ 0 (4.359564e-10) because we have a two-tailed alternative. Here T has a t distribution with 43 df.

c. The 95% confidence interval for the true mean difference is given by (if the samples are independent)

$(\bar{x}_{1}-\bar{x}_{2})\pm t_{0.05/2}s_{p}\sqrt{\frac{1}{25}+\frac{1}{20}}$ , i.e.,

$-0.03\pm t_{0.025}0.012459\sqrt{\frac{1}{25}+\frac{1}{20}}$

where $t_{0.025}$ is the 2.5th quantile of the t distribution with (25+20-2) = 43 degrees of freedom. So

$-0.03\pm(2.0167)(0.012459)(0.3)$ , i.e.,

(-0.0225, -0.0375)

8 0

2 years ago

A board game is played as follows: a number cube numbered 1 to 6 is rolled. If the number 1 comes up, you get nothing. If the nu

Evgesh-ka [11]

I think it is fair becasue you have the chance to win more than to lose point if get 1 or 6 you dont get nothing so that not a lose

4 0

2 years ago

A sample of an unknown chemical element naturally loses its mass over time. The relationship between the elapsed time ttt, in mo

Naddik [55]

Answer:

Step-by-step explanation:

Given a sample M(t)

M(t) = 120 • ( 81 / 625)^t

When is the fraction of the mass decay to 3/5 of it's mass

Generally

M(t) = Mo•(k^t)

The original mass is 120

Mo = 120

So, we want to find time when it decay to 3/5 of it's original mas

M = 3/5 × 120

M = 72

Then,

M(t) = 120 • ( 81 / 625)^t

72 = 120 • ( 81 / 625)^t

72 / 120 = ( 81 / 625)^t

0.6 = ( 81 / 625)^t

Take natural logarithmic of both sides

In(0.6) = In(81/625)^t

In(0.6) = t•In(81/625)

t = In(0.6) / In(81/625)

t = In(0.6) / In(0.1296)

t = 0.25 monthly

t = ¼ monthly

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2 years ago

A supermarket employee is making a mixture of cashews and almonds. Cashews cost $7 per pound, and almonds cost $5 per pound. The

svetoff [14.1K]

Because he needs less than (note that he didn't say he needs no more than) 6 pounds, first inequality is C) x+y<6

Now in inequalities that have 7 and 5 in them, 7 represents the cost of first ingredient and 5 the cost of second ingredient. If we multiply each of that number with quantity of that ingredient and sum both up we get the price of the mixture.

Because he says that he wants that mixture costs no more than 30 that means that it costs less than 30 but 30 can be the answer as well.

The last answer is E) 7x + 5y $\leq 30$

8 0

2 years ago