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3 years ago

HELP PLEASE ILL MARK BRAINLIEST

Mathematics

2 answers:

Paul [167]3 years ago

7 0

You answer would be b! hope this helps

gavmur [86]3 years ago

4 0

Answer:

B. $\frac{3+\sqrt{29} }{2}$

Step-by-step explanation:

$\frac{-(-3)\sqrt{(-3^2-4(1)(5)} }{(2)(1)}$

$=\frac{-(-3)+\sqrt{29} }{2}$

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Given: sinθ = -3/5, θ is a third quadrant angle, and tan φ = -7/24, φ is a second-quadrant angle; find cos(θ + φ])

Natali [406]

Answer:

First option: cos(θ + φ) = -117/125

Step-by-step explanation:

Recall that cos(θ + φ) = cos(θ)cos(φ) - sin(θ)sin(φ)

If sin(θ) = -3/5 in Quadrant III, then cos(θ) = -4/5.

Since tan(φ) = sin(φ)/cos(φ), then sin(φ) = -7/25 and cos(φ) = 24/25 in Quadrant II.

Therefore:

cos(θ + φ) = cos(θ)cos(φ) - sin(θ)sin(φ)

cos(θ + φ) = (-4/5)(24/25) - (-3/5)(-7/25)

cos(θ + φ) = (-96/125) - (21/125)

cos(θ + φ) = -96/125 - 21/125

cos(θ + φ) = -117/125

8 0

2 years ago

A survey of 1,562 randomly selected adults showed that 522 of them have heard of a new electronic reader. The accompanying techn

tester [92]

Answer:

a) We want to test the claim that 35% of adults have heard of the new electronic reader, then the system of hypothesis are.:

Null hypothesis: $p=0.35$

Alternative hypothesis: $p \neq 0.35$

And is a two tailed test

b) $z=\frac{0.334 -0.35}{\sqrt{\frac{0.35(1-0.35)}{1562}}}=-1.326$

c) $p_v =2*P(z$

d) Null hypothesis: $p=0.35$

e) Fail to reject the null hypothesis because the P-value is greater than the significance level, alpha.

Step-by-step explanation:

Information provided

n=1562 represent the random sample selected

X=522 represent the people who have heard of a new electronic reader

$\hat p=\frac{522}{1562}=0.334$ estimated proportion of people who have heard of a new electronic reader

$p_o=0.35$ is the value to verify

$\alpha=0.05$ represent the significance level

z would represent the statistic

$p_v$ represent the p value

Part a

We want to test the claim that 35% of adults have heard of the new electronic reader, then the system of hypothesis are.:

Null hypothesis: $p=0.35$

Alternative hypothesis: $p \neq 0.35$

And is a two tailed test

Part b

The statistic for this case is given :

$z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}}$ (1)

Replacing the info given we got:

$z=\frac{0.334 -0.35}{\sqrt{\frac{0.35(1-0.35)}{1562}}}=-1.326$

Part c

We can calculate the p value using the laternative hypothesis with the following probability:

$p_v =2*P(z$

Part d

The null hypothesis for this case would be:

Null hypothesis: $p=0.35$

Part e

The best conclusion for this case would be:

Fail to reject the null hypothesis because the P-value is greater than the significance level, alpha.

5 0

3 years ago