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5 + 0.55x = 10 + 0.75x

mylen [45]

Step-by-step explanation:

hope it is helpful to you

4 0

2 years ago

Read 2 more answers

“An architect is planning to incorporate several stone spheres of different sizes into the landscaping of a public park, and wor

Artist 52 [7]

Answer:

Part 1) The value that is closest to the cost of finishing a sphere with a 5.50-meter circumference is $900

Part 2) The value that is closest to the cost of finishing a sphere with a 7.85-meter circumference is $1,800

Step-by-step explanation:

Step 1

Find the radius of each sphere

we know that

The circumference of a circle is equal to

$C=2\pi r$

Find the radius of the sphere with a 5.50-meter circumference

For $C=5.50\ m$

assume

$\pi =3.14$

substitute and solve for r

$5.50=2(3.14)r$

$r=5.50/[2(3.14)]=0.88\ m$

Find the radius of the sphere with a 7.85-meter circumference

For $C=7.85\ m$

assume

$\pi =3.14$

substitute and solve for r

$7.85=2(3.14)r$

$r=7.85/[2(3.14)]=1.25\ m$

step 2

Find the surface area of each sphere

The surface area of sphere is equal to

$SA=4\pi r^{2}$

Find the surface area of sphere with a 5.50-meter circumference

For $r=0.88\ m$

assume

$\pi =3.14$

substitute

$SA=4(3.14)(0.88)^{2}$

$SA=9.73\ m^{2}$

Find the surface area of sphere with a 7.85-meter circumference

For $r=1.25\ m$

assume

$\pi =3.14$

substitute

$SA=4(3.14)(1.25)^{2}$

$SA=19.63\ m^{2}$

step 3

Find the cost of finishing each sphere

we know that

To find out the cost , multiply the surface area by $92 per square meter

Find the cost of sphere with a 5.50-meter circumference

$9.73*(92)=\$895.16$

therefore

The value that is closest to the cost of finishing a sphere with a 5.50-meter circumference is $900

Find the cost of sphere with a 7.85-meter circumference

$19.63*(92)=\$1,805.96$

therefore

The value that is closest to the cost of finishing a sphere with a 7.85-meter circumference is $1,800

6 0

3 years ago

35% of salt is in feta cheese. What percentage of salt is in that 350 g cheese?

xxMikexx [17]

The amount of salt is in the 350 g of feta cheese, when there is 35% of salt is in cheese, is 122.5 grams.

<h3>What is percentage of a number?</h3>

Percentage of a number is the part of the whole number which is expressed in the fraction of hundredth. It is represented with "%" symbol.

35% of salt is in feta cheese. The percentage of salt is in that 350 g cheese has to be found out.

Let suppose there is x grams of slat in feta cheese of 350 g. Thus, the amount of salt (35% of 350) is,

$x=\dfrac{35}{100}\times350\\x=122.5\rm\; gm$

Thus, the amount of salt is in the 350 g of feta cheese, when there is 35% of salt is in cheese, is 122.5 grams.

Learn more about the percentage here;

brainly.com/question/2085058

#SPJ1

5 0

1 year ago

Please help me with this MATH question.

Pie

Answer:

g= -2

Step-by-step explanation:

3 0

3 years ago

Read 2 more answers

Suppose a geyser has a mean time between eruptions of 72 minutes. Let the interval of time between the eruptions be normally dis

nikitadnepr [17]

Answer:

(a) The probability that a randomly selected time interval between eruptions is longer than 82 minutes is 0.3336.

(b) The probability that a random sample of 13-time intervals between eruptions has a mean longer than 82 minutes is 0.0582.

(c) The probability that a random sample of 34 time intervals between eruptions has a mean longer than 82 minutes is 0.0055.

(d) Due to an increase in the sample size, the probability that the sample mean of the time between eruptions is greater than 82 minutes decreases because the variability in the sample mean decreases as the sample size increases.

(e) The population mean must be more than 72, since the probability is so low.

Step-by-step explanation:

We are given that a geyser has a mean time between eruptions of 72 minutes.

Also, the interval of time between the eruptions be normally distributed with a standard deviation of 23 minutes.

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

So, X ~ N( $\mu=72, \sigma^{2} =23^{2}$ )

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 = 72 minutes

$\sigma$ = standard deviation = 23 minutes

Now, the probability that a randomly selected time interval between eruptions is longer than 82 minutes is given by = P(X > 82 min)

P(X > 82 min) = P( $\frac{X-\mu}{\sigma}$ > $\frac{82-72}{23}$ ) = P(Z > 0.43) = 1 - P(Z $\leq$ 0.43)

= 1 - 0.6664 = 0.3336

The above probability is calculated by looking at the value of x = 0.43 in the z table which has an area of 0.6664.

(b) Let $\bar X$ = sample mean time between the eruptions

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 = 72 minutes

$\sigma$ = standard deviation = 23 minutes

n = sample of time intervals = 13

Now, the probability that a random sample of 13 time intervals between eruptions has a mean longer than 82 minutes is given by = P( $\bar X$ > 82 min)

P( $\bar X$ > 82 min) = P( $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ > $\frac{82-72}{\frac{23}{\sqrt{13} } }$ ) = P(Z > 1.57) = 1 - P(Z $\leq$ 1.57)

= 1 - 0.9418 = 0.0582

The above probability is calculated by looking at the value of x = 1.57 in the z table which has an area of 0.9418.

(c) Let $\bar X$ = sample mean time between the eruptions

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 = 72 minutes

$\sigma$ = standard deviation = 23 minutes

n = sample of time intervals = 34

Now, the probability that a random sample of 34 time intervals between eruptions has a mean longer than 82 minutes is given by = P( $\bar X$ > 82 min)

P( $\bar X$ > 82 min) = P( $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ > $\frac{82-72}{\frac{23}{\sqrt{34} } }$ ) = P(Z > 2.54) = 1 - P(Z $\leq$ 2.54)

= 1 - 0.9945 = 0.0055

The above probability is calculated by looking at the value of x = 2.54 in the z table which has an area of 0.9945.

(d) Due to an increase in the sample size, the probability that the sample mean of the time between eruptions is greater than 82 minutes decreases because the variability in the sample mean decreases as the sample size increases.

(e) If a random sample of 34-time intervals between eruptions has a mean longer than 82 minutes, then we conclude that the population mean must be more than 72, since the probability is so low.

6 0

3 years ago