Can someone please help to write the standard form of this hyperbola?

At the farmers market a farmer has bunches of radishes for sale. The

stealth61 [152]

Answer:

The probability a random selected radish bunch weighs between 5 and 6.5 ounces is 0.8185

Step-by-step explanation:

The weight of the radish bunches is normally distributed with a mean of 6 ounces and a standard deviation of 0.5 ounces

Mean = $\mu = 6$

Standard deviation = $\sigma = 0.5$

We are supposed to find the probability a random selected radish bunch weighs between 5 and 6.5 ounces i.e.P(5<x<6.5)

$Z=\frac{x-\mu}{\sigma}$

At x = 5

$Z=\frac{5-6}{0.5}$

Z=-2

$Z=\frac{x-\mu}{\sigma}$

At x = 6.5

$Z=\frac{6.5-6}{0.5}$

Z=1

Refer the z table for p value

P(5<x<6.5)=P(x<6.5)-P(x<5)=P(Z<1)-P(Z<-2)=0.8413-0.0228=0.8185

Hence the probability a random selected radish bunch weighs between 5 and 6.5 ounces is 0.8185

6 0

3 years ago

6. Three bunches of flowers can be bought for $9.48. Four bunches can be bought for $10.64. Which is the better buy?

lawyer [7]

The better buy would be four bunches for $10.66 because each bunch of flowers would be $2.66 compared to $3.16 per bunch.

5 0

3 years ago

A 1,200g block of phosphorus-32, which has a half life of 14.3 days, is stored for 100.1 days. At the end of this period, how mu

Valentin [98]

Answer:

The amount of phosphorus-32 left after 100.1 days is 9.3 g.

Step-by-step explanation:

Given:

Initial amount of Phosphorus-32 is, $N_0=1200\ g$

Time period of decay is, $t=100.1\ days$

Half life of the block is, $t_{1/2}=14.3\ days$

Now, final amount left is, $N=?$

We know that, the decay equation for a radioactive material is given as:

$N=N_0e^{-kt}\\k\to decay\ constant$

The value of the decay constant is given as:

$k=\frac{\ln 2}{t_{1/2}}\\\\k=\frac{0.693}{14.3}=0.0485$

Now, plug in all the given values and calculate 'N'. This gives,

$N=(1200)e^{(-0.0485\times 100.1)}\\\\N=9.349\approx 9.3\ g$

Therefore, the amount of phosphorus-32 left after 100.1 days is 9.3 g

5 0

3 years ago

A sample of n = 4 scores is obtained from a population with a mean of 70 and a standard deviation of 8. If the sample mean corre

jeka57 [31]

Answer:

The value of the sample mean is 78.

Step-by-step explanation:

We are given that a sample of n = 4 scores is obtained from a population with a mean of 70 and a standard deviation of 8.

Also, the sample mean corresponds to a z score of 2.00.

Let $\bar X$ = sample mean

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

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

where, $\mu$ = population mean = 70

$\sigma$ = standard deviation = 8

n = sample size = 4

The Z-score measures how many standard deviations the measure is away from the mean. After finding the Z-score, we look at the z-score table and find the p-value (area) associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X.

Now, we are given that the sample mean corresponds to a z score of 2.00 for which we have to find the value of sample mean;

So, z-score formula is given by ;

z-score = $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ = $\frac{\bar X-70}{\frac{8}{\sqrt{4} } }$

2.00 = $\frac{\bar X-70}{\frac{8}{\sqrt{4} } }$

2.00 = $\frac{\bar X-70}{4 } }$

$\bar X = 70+(2 \times 4)$

$\bar X$ = 70 + 8 = 78

Therefore, the value of the sample mean is 78.

3 0

3 years ago

You have 275 square inches of wrapping paper. Do you have enough wrapping paper to wrap the gift box shown?

Zigmanuir [339]

The surface area of the box is bigger than the surface area of the available wrapping paper. Thus, you don't have enough wrapping paper to wrap the gift box.

The given parameters;

area of the wrapping paper available, A = 275 square inches
volume of the gift box, V = 343 in²

First, determine the length of each side of the box;

The length of each side is represented with x;

x³ = 343

$x = \sqrt[3]{343} \\\\x = 7 \ inches$

Second, find the surface area of the box;

surface area of the box, A = 6x²

A = 6(7)²

A = 294 in²

The surface area of the box is bigger than the surface area of the available wrapping paper. Thus, you don't have enough wrapping paper to wrap the gift box.

Learn more here: brainly.com/question/15908008

8 0

3 years ago