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3 years ago

Https://brainly.com/question/12057829 CAN SOMEONE TELL ME HOW TO GRAPH THIS

Mathematics

2 answers:

olga55 [171]3 years ago

6 0

we can draw the graph of cosine with amplitude 2.

Step-by-step explanation:

In the given function g(x) = 2cosx we know the following parameters which are helpful to draw the graph.

Its a cosine graph given in the form of y = acos(Bx + c) + d

Here a is amplitude, period = , c = phase shift, d = vertical shift

We know a = 2, B = , c = 0, d = 0.

GenaCL600 [577]3 years ago

6 0

Answer:

You graph it but

Step-by-step explanation:

Ghyyyyyy ya know it’s nottttt harddd lollipops

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Y (y-4) – (y-2) (y-3)=10

Alika [10]

Answer:

$(y {}^{2} - 4y) - ( {y}^{2} - 3y - 2y - 6) \\ - 4y - 3y - 2y - 6 \\ - 9y - 6 \: equals \: 10 \\ - 9y \: equals16 \\ y \: equals \: \frac{16}{ - 9}$

y = -(16/9)

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3 years ago

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lorasvet [3.4K]

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<h2>D 5 This the answer </h2>

Step-by-step explanation:

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2 years ago

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A student who is training for a distance race jogged for thirty minutes and covered a distance of 7.5 km. What is the average sp

Tju [1.3M]

Answer:

<h2>Average speed is 15km/h</h2>

Step-by-step explanation:

Step one:

given data:

total distance =7.5km

time taken= 30 min= 0.5 hours

we know that the average speed is the average distance divided by time taken

Step two:

mathematically expressed as

speed=distance/time

substituting we have

speed=7.5/0.5

speed=15km/hour

Hence the average speed is 15km/h

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3 years ago

If x+y=z and z=x+4,then x+y=x+4

lidiya [134]

That is true because as you can see in the other 2 equations, they both have z on one of the sides meaning that the sides that don't have z are equal to each other

Hope this helps :)

3 0

3 years ago

The amount of syrup that people put on their pancakes is normally distributed with mean 63 mL and standard deviation 13 mL. Supp

andreyandreev [35.5K]

Answer:

(a) X ~ N( $\mu=63, \sigma^{2} = 13^{2}$ ).

$\bar X$ ~ N( $\mu=63,s^{2} = (\frac{13}{\sqrt{43} } )^{2}$ ).

(b) If a single randomly selected individual is observed, the probability that this person consumes is between 61.4 mL and 62.8 mL is 0.0398.

(c) For the group of 43 pancake eaters, the probability that the average amount of syrup is between 61.4 mL and 62.8 mL is 0.2512.

(d) Yes, for part (d), the assumption that the distribution is normally distributed necessary.

Step-by-step explanation:

We are given that the amount of syrup that people put on their pancakes is normally distributed with mean 63 mL and a standard deviation of 13 mL.

Suppose that 43 randomly selected people are observed pouring syrup on their pancakes.

(a) Let X = <u><em>amount of syrup that people put on their pancakes</em></u>

The z-score probability distribution for the normal distribution is given by;

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

where, $\mu$ = mean amount of syrup = 63 mL

$\sigma$ = standard deviation = 13 mL

So, the distribution of X ~ N( $\mu=63, \sigma^{2} = 13^{2}$ ).

Let $\bar X$ = <u><em>sample mean amount of syrup that people put on their pancakes</em></u>

The z-score probability distribution for the sample mean is given by;

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

where, $\mu$ = mean amount of syrup = 63 mL

$\sigma$ = standard deviation = 13 mL

n = sample of people = 43

So, the distribution of $\bar X$ ~ N( $\mu=63,s^{2} = (\frac{13}{\sqrt{43} } )^{2}$ ).

(b) If a single randomly selected individual is observed, the probability that this person consumes is between 61.4 mL and 62.8 mL is given by = P(61.4 mL < X < 62.8 mL)

P(61.4 mL < X < 62.8 mL) = P(X < 62.8 mL) - P(X $\leq$ 61.4 mL)

P(X < 62.8 mL) = P( $\frac{X-\mu}{\sigma}$ < $\frac{62.8-63}{13}$ ) = P(Z < -0.02) = 1 - P(Z $\leq$ 0.02)

= 1 - 0.50798 = 0.49202

P(X $\leq$ 61.4 mL) = P( $\frac{X-\mu}{\sigma}$ $\leq$ $\frac{61.4-63}{13}$ ) = P(Z $\leq$ -0.12) = 1 - P(Z < 0.12)

= 1 - 0.54776 = 0.45224

Therefore, P(61.4 mL < X < 62.8 mL) = 0.49202 - 0.45224 = 0.0398.

(c) For the group of 43 pancake eaters, the probability that the average amount of syrup is between 61.4 mL and 62.8 mL is given by = P(61.4 mL < $\bar X$ < 62.8 mL)

P(61.4 mL < $\bar X$ < 62.8 mL) = P( $\bar X$ < 62.8 mL) - P( $\bar X$ $\leq$ 61.4 mL)

P( $\bar X$ < 62.8 mL) = P( $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ < $\frac{62.8-63}{\frac{13}{\sqrt{43} } }$ ) = P(Z < -0.10) = 1 - P(Z $\leq$ 0.10)

= 1 - 0.53983 = 0.46017

P( $\bar X$ $\leq$ 61.4 mL) = P( $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ $\leq$ $\frac{61.4-63}{\frac{13}{\sqrt{43} } }$ ) = P(Z $\leq$ -0.81) = 1 - P(Z < 0.81)

= 1 - 0.79103 = 0.20897

Therefore, P(61.4 mL < X < 62.8 mL) = 0.46017 - 0.20897 = 0.2512.

(d) Yes, for part (d), the assumption that the distribution is normally distributed necessary.

4 0

3 years ago