What's the answer this is equivalent fractions

I need help fast like rn

lbvjy [14]

Answer:

vyuWYIRTG3ñwyiehg

Step-by-step explanation:

4 0

3 years ago

joel has 24 sports trophies. of the trophies, 1/8 are soccer trophies. how many soccer trophies does joel have?

hichkok12 [17]

Answer:

3 soccer trophies does Joel have

Step-by-step explanation:

As per the statement:

Joel has 24 sports trophies.

⇒Total sport trophies Joel have= 24

It is also given: of the trophies, 1/8 are soccer trophies

We have to find how many soccer trophies does Joel have.

$\text{Joel have soccer trophies} =\frac{1}{8} \cdot 24$

Simplify:

$\text{Joel have soccer trophies} =3$

Therefore, 3 soccer trophies does Joel have

3 0

3 years ago

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Triangle ABC has vertices at A(−3, 4), B(4, −2), C(8, 3). The triangle translates 2 units up and 1 unit right. Which rule repres

Nataly [62]

Answer:

The translation statement is given by:

$(x,y)\rightarrow (x+1,y+2)$

After the translation, the coordinates of vertex A is (-2,6).

Step-by-step explanation:

Given :

Vertices of a triangle ABC are:

A(−3, 4), B(4, −2), C(8, 3)

The triangle is translated 2 units up and 1 unit right.

To find the co-ordinates of point A after translation.

Translation rules.

For shift of $c$ units up, the translation is given as:

$(x,y)\rightarrow (x,y+c)$

For shift of $k$ units right, the translation is given as:

$(x,y)\rightarrow (x+k,y)$

So, it says the triangles is translated 2 units up and 1 unit right.

The translation statement is given by:

$(x,y)\rightarrow (x+1,y+2)$

So, co-ordinates of point A after translation is given by :

$(-3,4)\rightarrow (-3+1,4+2)=(-2,6)$

After the translation, the coordinates of vertex A is (-2,6).

4 0

3 years ago

Suppose a geyser has a mean time between irruption’s of 75 minutes. If the interval of time between the eruption is normally dis

lesya [120]

Answer:

(a) The probability that a randomly selected Time interval between irruption is longer than 84 minutes is 0.3264.

(b) The probability that a random sample of 13 time intervals between irruption has a mean longer than 84 minutes is 0.0526.

(c) The probability that a random sample of 20 time intervals between irruption has a mean longer than 84 minutes is 0.0222.

(d) The probability decreases because the variability in the sample mean decreases as we increase the sample size

(e) The population mean may be larger than 75 minutes between irruption.

Step-by-step explanation:

We are given that a geyser has a mean time between irruption of 75 minutes. Also, the interval of time between the eruption is normally distributed with a standard deviation of 20 minutes.

(a) Let X = the interval of time between the eruption

So, X ~ Normal( $\mu=75, \sigma^{2} =20$ )

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

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

where, $\mu$ = population mean time between irruption = 75 minutes

$\sigma$ = standard deviation = 20 minutes

Now, the probability that a randomly selected Time interval between irruption is longer than 84 minutes is given by = P(X > 84 min)

P(X > 84 min) = P( $\frac{X-\mu}{\sigma}$ > $\frac{84-75}{20}$ ) = P(Z > 0.45) = 1 - P(Z $\leq$ 0.45)

= 1 - 0.6736 = 0.3264

The above probability is calculated by looking at the value of x = 0.45 in the z table which has an area of 0.6736.

(b) Let $\bar X$ = sample time intervals between the eruption

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, $\mu$ = population mean time between irruption = 75 minutes

$\sigma$ = standard deviation = 20 minutes

n = sample of time intervals = 13

Now, the probability that a random sample of 13 time intervals between irruption has a mean longer than 84 minutes is given by = P( $\bar X$ > 84 min)

P( $\bar X$ > 84 min) = P( $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ > $\frac{84-75}{\frac{20}{\sqrt{13} } }$ ) = P(Z > 1.62) = 1 - P(Z $\leq$ 1.62)

= 1 - 0.9474 = 0.0526

The above probability is calculated by looking at the value of x = 1.62 in the z table which has an area of 0.9474.

(c) Let $\bar X$ = sample time intervals between the eruption

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, $\mu$ = population mean time between irruption = 75 minutes

$\sigma$ = standard deviation = 20 minutes

n = sample of time intervals = 20

Now, the probability that a random sample of 20 time intervals between irruption has a mean longer than 84 minutes is given by = P( $\bar X$ > 84 min)

P( $\bar X$ > 84 min) = P( $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ > $\frac{84-75}{\frac{20}{\sqrt{20} } }$ ) = P(Z > 2.01) = 1 - P(Z $\leq$ 2.01)

= 1 - 0.9778 = 0.0222

The above probability is calculated by looking at the value of x = 2.01 in the z table which has an area of 0.9778.

(d) When increasing the sample size, the probability decreases because the variability in the sample mean decreases as we increase the sample size which we can clearly see in part (b) and (c) of the question.

(e) Since it is clear that the probability that a random sample of 20 time intervals between irruption has a mean longer than 84 minutes is very slow(less than 5%0 which means that this is an unusual event. So, we can conclude that the population mean may be larger than 75 minutes between irruption.

8 0

3 years ago

Please help will mark brainliest!

dmitriy555 [2]

2x - 3y = - 2

3x - 2y = 12

the value of X in the solution to the system is 4.

6 0

3 years ago

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