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Nitella [24]
3 years ago
14

If sin=2/3 and tan is less than 0, what is the value of cos

Mathematics
1 answer:
Sonbull [250]3 years ago
4 0

tangent is less than 0 or tan(θ) < 0, is another way to say tan(θ) is negative, well, that only happens on the II Quadrant and IV Quadrant, where sine and cosine are different signs, so we know θ is on the II or IV Quadrant.

\bf sin(\theta )=\cfrac{\stackrel{opposite}{2}}{\stackrel{hypotenuse}{3}}\qquad \impliedby \textit{let's find the \underline{adjacent side}} \\\\\\ \textit{using the pythagorean theorem} \\\\ c^2=a^2+b^2\implies \pm\sqrt{c^2-b^2}=a \qquad \begin{cases} c=hypotenuse\\ a=adjacent\\ b=opposite\\ \end{cases}

\bf \pm\sqrt{3^2-2^2}=a\implies \pm\sqrt{5}=a\implies \stackrel{\textit{II Quadrant}}{-\sqrt{5}=a} \\\\[-0.35em] ~\dotfill\\\\ ~\hfill cos(\theta )=\cfrac{\stackrel{adjacent}{-\sqrt{5}}}{\stackrel{hypotenuse}{3}}~\hfill

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Step-by-step explanation:

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2 years ago
The distribution of weights for newborn babies is approximately normally distributed with a mean of 7.4 pounds and a standard de
blsea [12.9K]

Answer:

1. 15.87%

2.  6 pounds and 8.8 pounds.

3. 2.28%

4. 50% of newborn babies weigh more than 7.4 pounds.

5. 84%

Step-by-step explanation:

We are given the following information in the question:

Mean, μ = 7.4 pounds

Standard Deviation, σ = 0.7 pounds

We are given that the distribution of weights for newborn babies is a bell shaped distribution that is a normal distribution.

Formula:

z_{score} = \displaystyle\frac{x-\mu}{\sigma}

1.Percent of newborn babies weigh more than 8.1 pounds

P(x > 8.1)

P( x > 8.1) = P( z > \displaystyle\frac{8.1 - 7.4}{0.7}) = P(z > 1)

= 1 - P(z \leq 1)

Calculation the value from standard normal z table, we have,  

P(x > 8.1) = 1 - 0.8413 = 0.1587 = 15.87\%

15.87% of newborn babies weigh more than 8.1 pounds.

2.The middle 95% of newborn babies weight

Empirical Formula:

  • Almost all the data lies within three standard deviation from the mean for a normally distributed data.
  • About 68% of data lies within one standard deviation from the mean.
  • About 95% of data lies within two standard deviations of the mean.
  • About 99.7% of data lies within three standard deviation of the mean.

Thus, from empirical formula 95% of newborn babies will lie between

\mu-2\sigma= 7.4-2(0.7) = 6\\\mu+2\sigma= 7.4+2(0.7)=8.8

95% of newborn babies will lie between 6 pounds and 8.8 pounds.

3. Percent of newborn babies weigh less than 6 pounds

P(x < 6)

P( x < 6) = P( z > \displaystyle\frac{6 - 7.4}{0.7}) = P(z < -2)

Calculation the value from standard normal z table, we have,  

P(x < 6) =0.0228 = 2.28\%

2.28% of newborn babies weigh less than 6 pounds.

4. 50% of newborn babies weigh more than pounds.

The normal distribution is symmetrical about mean. That is the mean value divide the data in exactly two parts.

Thus, approximately 50% of newborn babies weigh more than 7.4 pounds.

5. Percent of newborn babies weigh between 6.7 and 9.5 pounds

P(6.7 \leq x \leq 9.5)\\\\ = P(\displaystyle\frac{6.7 - 7.4}{0.7} \leq z \leq \displaystyle\frac{9.5-7.4}{0.7})\\\\ = P(-1 \leq z \leq 3)\\\\= P(z \leq 3) - P(z < -1)\\= 0.9987 -0.1587= 0.84 = 84\%

84% of newborn babies weigh between 6.7 and 9.5 pounds.

7 0
3 years ago
I need help with this!
Viefleur [7K]

Answer:

the answer is option E.

Step-by-step explanation:

it is in the form f/g (x).

we know f and g from the equation;

we can rewrite it as, (√(9-x^2)/(3x-1))

if you notice you cannot put any number less than -3 or greater than 3 in the numeror because if you do you get a negative root which is false. for instance if you put 4 or -4 in the numerator you get 9 - (4 or -4 square ) which is 9- 16 which is a negative number and you cannot take root of a negative number.

on the numerator if you put 1/3 as the value for x you will get zero in the denominator. and any number divided by zero is undefined so that cannot be.

this means that option E is the right one that satisfies the condition. it means the domain is [-3, 1/3) U (1/3, 3] .

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