Gareth buys two oranges. He pays with a £1 coin and gets 52p change. Work out the cost of one orange.

Find the area of the following shape. You must show all work to receive credit.

SVEN [57.7K]

Answer:

Step-by-step explanation:

A = [(3x2)/2] + [(4x2)/2] - 1/2

A = 3 + 4 - 1/2

A = 6.5

3 0

3 years ago

Check each equation whose graph is the line that

yanalaym [24]

Answer:

Points P ( 4 , - 7 ) and Q ( 1 , 5 ) belong to the equations:

1 ; 4 ; and 5

Step-by-step explanation:

Equations 1 ; 4 and 5 are the same equation

Equation 1 y = - 4*x + 9

Equation 4

y + 7 = - 4 * ( x - 4 ) ⇒ y + 7 = - 4*x + 16 ⇒ y = - 4*x - 7 + 16

y = -4*x + 9

Equation 5

4*x + y = 9 ⇒ y = - 4*x + 9

Now for the equation y = - 4*x + 9

P ( 4 , -7)

For x = 4 y = - 4*(4) + 9 ⇒ y = - 16 + 9 ⇒ y = - 7

Then point P is in the line y = - 4*x + 9

Point Q (1 , 5 )

For x = 1 y = - 4 * ( 1) + 9 ⇒ y = - 4 + 9 ⇒ y = 5

Point Q is in the line y = - 4*x + 9

Equation 2

y = - 4*x - 23

Point P ( 4 , - 7 )

For x = 4 y = 16 - 23 y = - 7

Point P is in the line

Point Q

For x = 1 y = - 4 *(1) - 23 ⇒ y = - 27

Then poin Q is not in the line

Equation 3

y - 1 = - 4 * ( x - 4 )

y - 1 = -4*x + 16 ⇒ y = - 4*x + 17

Point P ( 4 , - 7 )

For x = 4

y = - 16 + 17 ⇒ y = 1

Point P is not in the line

And Point Q ( 1 , 5 )

For x = 1

y = - 4* ( 1 ) + 17 ⇒ y = 13 Q is not in the line

3 0

3 years ago

n Monday 2⁄3 of the team practiced, Tuesday 7⁄8 , Wednesday 1⁄2 , and Thursday 3⁄4 . On which day were the most team members pre

Tanya [424]

Answer is Tuesday because if you turn all the fractions into a percentage then Tuesday has the highest percentage.

4 0

3 years ago

Read 2 more answers

Consider a steel guitar string of initial length L=1.00 meter and cross-sectional area A=0.500 square millimeters. The Young's m

Goshia [24]

Answer:

15mm

Step-by-step explanation:

we are looking for extension

To get Extension you need the original length and the strain both of which you are given

initial length L = 1.00m

the area A = 0.5mm² = 0.5 mm² = 0.5 x 10⁻⁶ m² ( we are changing to metres squared)

E = 2.0 x 10¹¹ n/m², Young's modulus

P = 1500N, the applied tension

Now to Calculate the stress.

σ = P/A (force/area) = (1500 N)/(0.5 x 10⁻⁶ m²) = 3 x 10⁹ N/m²

Also, Let β = the stretch of the string.

Then the strain is

ε = β/L (extension/ original length)

By definition, the strain is ε = σ/E = (3 x 10⁹ N/m²)/(2 x 10¹¹ N/m²) = 0.015

Therefore β/(1 m) = 0.015β = 0.015 m = 15 mm

Answer: 15 mm

8 0

3 years ago

The time for a visitor to read health instructions on a Web site is approximately normally distributed with a mean of 10 minutes

klio [65]

Answer:

a) The mean is 10 and the variance is 0.0625.

b) 0.6826 = 68.26% probability that the mean time of the visitors is within 15 seconds of 10 minutes.

c) 10.58 minutes.

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal Probability Distribution

Problems of normal distributions can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the z-score of a measure X is given by:

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

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value 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. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem establishes that, for a normally distributed random variable X, with mean $\mu$ and standard deviation $\sigma$ , the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$ .

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

Normally distributed with a mean of 10 minutes and a standard deviation of 2 minutes.

This means that $\mu = 10, \sigma = 2$

Suppose 64 visitors independently view the site.

This means that $n = 64, = \frac{2}{\sqrt{64}} = 0.25$

a. The expected value and the variance of the mean time of the visitors.

Using the Central Limit Theorem, mean of 10 and variance of (0.25)^2 = 0.0625.

b. The probability that the mean time of the visitors is within 15 seconds of 10 minutes.

15 seconds = 15/60 = 0.25 minutes, so between 9.75 and 10.25 seconds, which is the p-value of Z when X = 10.25 subtracted by the p-value of Z when X = 9.75.

X = 10.25

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

By the Central Limit Theorem

$Z = \frac{X - \mu}{s}$