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2 years ago

8

Which equation represents the graph of Charli's line? A. y= -3/4x-3. B. Y= 3/4x- 4. C. y = -4/3x- 4. D. y= 4/3x- 3. LOOK AT THE

PHOTO PLZZ

2 answers:

enot [183]2 years ago

5 0

Saya rase jwpn di C . Kalau bukan minx maaf lah ye

kykrilka [37]2 years ago

4 0

C. i’m pretty sure because it goes up by 4 and over 3. And the intercept is 4

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the circumference of a circle is a(pi) units, and the area of the circle is b(pi) square units. If a=b, what is the radius of th

xxTIMURxx [149]

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the redius is 0 unit or 1/2^(1/2) unit.

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3 years ago

3 pounds of gummie bears cost $8.97. <br> What will one pound cost?

KATRIN_1 [288]

Answer:

2.99

Step-by-step explanation:

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2 years ago

TRIGONOMETRY , can you solve this plz:(<br>even if you know one of them,plz help:)

daser333 [38]

Answer:

See below

Step-by-step explanation:

$\frac{1}{1-cos A} - \frac{1}{1+cos A} =2 cosec A cot A \\ \\ L. H. S.= \frac{1}{1-cos A} - \frac{1}{1+cos A} \\ \\ = \frac{1+cos A - (1 - cos A)}{(1-cos A)(1+cos A)}\\ \\ = \frac{1+cos A - 1 + cos A}{1-cos^{2} A}\\ \\ = \frac{2cos A }{sin^{2} A}\\ \\ =2 .\frac{1}{sin A}.\frac{cos A }{sin A}.\\ \\ = 2.cosec \: A.cot \: A \\ \\ = R.H.S$

5 0

3 years ago

Solve:<br><img src="https://tex.z-dn.net/?f=%20%5Cfrac%7Bzz%7D%7B4%7D%20%20%2B%205%20%5Cgeqslant%205.5" id="TexFormula1" title="

allochka39001 [22]

$\bf \cfrac{zz}{4} + 5\geqslant 5.5\implies \stackrel{\textit{subtracting 5 from both sides}}{\cfrac{zz}{4}\geqslant 0.5}\implies \stackrel{\textit{multiplying both sides by 4}}{zz \geqslant 2}$

7 0

3 years ago

The score of golfers for a particular course follows a normal distribution that has a mean of 73 and a standard deviation of 3.

Artemon [7]

Answer:

P(X $\geq$ 74) = 0.3707

Step-by-step explanation:

We are given that the score of golfers for a particular course follows a normal distribution that has a mean of 73 and a standard deviation of 3.

Let X = Score of golfers

So, X ~ N( $\mu=73,\sigma^{2}=3^{2}$ )

The z score probability distribution is given by;

Z = $\frac{X-\mu}{\sigma}$ ~ N(0,1)

where, $\mu$ = population mean = 73

$\sigma$ = standard deviation = 3

So, the probability that the score of golfer is at least 74 is given by = P(X $\geq$ 74)

P(X $\geq$ 74) = P( $\frac{X-\mu}{\sigma}$ $\geq$ $\frac{74-73}{3}$ ) = P(Z $\geq$ 0.33) = 1 - P(Z < 0.33)

= 1 - 0.62930 = 0.3707

Therefore, the probability that the score of golfer is at least 74 is 0.3707 .

3 0

3 years ago