All categories

3 years ago

Check all that apply. If sec=13/12, then:

Mathematics

2 answers:

Verizon [17]3 years ago

3 0

<u>Answer</u>
1) TanФ = 5/12
2) CosФ = 12/13

<u>Explanation</u>
SecФ = 1/CosФ

Cos = adjacent/hypotenuse
SecФ= 13/12
cosФ = 12/13

This means that the adjacent is 12 and hypotenuse is 13.
Opposite = √(13²-12²)
= √25
= 5

Tangent = opposite/adjacent
TanФ = 5/12

Sine = opposite/hypotenuse
SinФ = 5/13

Nadya [2.5K]3 years ago

3 0

Answer:

1) TanФ = 5/12

2) CosФ = 12/13

Step-by-step explanation:

You might be interested in

HELP WILL MAKE BRAINLIST TO WHOEVER DID IN EXPLAINATION

Nataly_w [17]

Answer:

The scouts hiked up the mountain in 1 mile.

Step-by-step explanation:

7 0

3 years ago

Need help on number 25!

zhuklara [117]

Answer:

30000

Step-by-step explanation:

when dividing the powers, you subtract them.

8 0

3 years ago

Help me with this pls

SVEN [57.7K]

The answer is (A) 4 wfheyjtehetjtwy

5 0

3 years ago

Read 2 more answers

Can you help me with these two questions please !!!!

zubka84 [21]

Answer: $(x+2)^{2}+(y-4)^{2} =16$

and

$(x-6)^{2} +(y+4)^{2} =20$

Step-by-step explanation:

Equation for a Circle

$(x-h)^{2} +(y-k)^{2} =r^{2}$

h= -2 k= 4 r= 4

$(x+2)^{2}+(y-4)^{2} =16$

Part 2

Complete the Square

$x^{2} +y^{2} -12x+8y+32=0\\(x^{2}-12x+-)+(y^{2}+8y)=-32\\(x^{2} -12x+36)=(y^{2} +8y+16)=-32+36+16\\\\(x-6)^{2} +(y+4)^{2} =20$

6 0

3 years ago

Read 2 more answers

Suppose that the mean water hardness of lakes in Kansas is 425 mg/L and these values tend to follow a normal distribution. A lim

tino4ka555 [31]

Answer:

The mean water hardness of lakes in Kansas is 425 mg/L or greater.

Step-by-step explanation:

We are given the following data set:

346, 496, 352, 378, 315, 420, 485, 446, 479, 422, 494, 289, 436, 516, 615, 491, 360, 385, 500, 558, 381, 303, 434, 562, 496

Formula:

$\text{Standard Deviation} = \sqrt{\displaystyle\frac{\sum (x_i -\bar{x})^2}{n-1}}$

where $x_i$ are data points, $\bar{x}$ is the mean and n is the number of observations.

$Mean = \displaystyle\frac{\text{Sum of all observations}}{\text{Total number of observation}}$

$Mean =\displaystyle\frac{10959}{25} =438.36$

Sum of squares of differences = 175413.76

$S.D = \sqrt{\dfrac{175413.76}{24}} = 85.49$

Population mean, μ = 425 mg/L

Sample mean, $\bar{x}$ = 438.36

Sample size, n = 25

Alpha, α = 0.05

Sample standard deviation, s = 85.49

First, we design the null and the alternate hypothesis

$H_{0}: \mu \geq 425\text{ mg per Litre}\\H_A: \mu < 425\text{ mg per Litre}$

We use one-tailed t test to perform this hypothesis.

Formula:

$t_{stat} = \displaystyle\frac{\bar{x} - \mu}{\frac{s}{\sqrt{n}} }$

Putting all the values, we have

$t_{stat} = \displaystyle\frac{438.36 - 425}{\frac{85.49}{\sqrt{25}} } = 0.7813$

Now, $t_{critical} \text{ at 0.05 level of significance, 24 degree of freedom } = -1.7108$

Since,

The calculated t-statistic is greater than the critical value, we fail to reject the null hypothesis and accept it.

Thus, the mean water hardness of lakes in Kansas is 425 mg/L or greater.

6 0

3 years ago