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3 years ago

12

Keisha is walking around a track that is 400 yards long. How many more yards does she need to walk around the track to complete

2 miles ?

1 answer:

lutik1710 [3]3 years ago

6 0

400x4 is 1 mile so 400x8 is 2 miles so 3,200 is 2 miles minus 400 because she already walked that so the answer is 2,800

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A sample of 18 small bags of the same brand of candies was selected. Assume that the population distribution of bag weights is n

ki77a [65]

Answer:

a)

i. x=- - - - - -

ii. o=- - - - - -

iii. Sx=- - - - -

b) The random variable X signifies the weights of the bags of candies which are selected at random.

c) The statistics variable X is a measure on a sample that is used as an estimate of the population mean.X is the mean weight of the 16 bags of candies that were selected.

d) The distribution needed for this problem is normal distribution with parameters because the population standard deviation is known.

N(X, o/ $\sqrt{n}$ )

So, the distribution is

N(2, 0.12/ $\sqrt{16}$ )

e)

i. The 90% confidence interval for the population mean weight of the candies is 1.9589 , 2.0411

iii. The error bound is 0.0411

f)

i. The 98% confidence interval for the population mean weight of the candies is 1.9418 , 2.0582

iii. The error bound is 0.0582

g) The difference in the confidence intervals in part (f) and part (e) is of the level of confidence. The change is, the change in the area being calculated for the normal distribution. Therefore, the larger confidence level results in larger area and larger interval.

Hence the interval in part is larger than the one in part (e).

h) The interval in part (f) signifies that with 98% confidence that the population mean weight of the bag of candies lies between 1.942 ounce and 2.058 ounce.

Step-by-step explanation:

a)

i. x=- - - - - -

ii. o=- - - - - -

iii. Sx=- - - - -

b) The random variable X signifies the weights of the bags of candies which are selected at random.

c) The statistics variable X is a measure on a sample that is used as an estimate of the population mean.X is the mean weight of the 16 bags of candies that were selected.

d) The distribution needed for this problem is normal distribution with parameters because the population standard deviation is known.

N(X, o/ $\sqrt{n}$ )

So, the distribution is

N(2, 0.12/ $\sqrt{16}$ )

e)

i. The 90% confidence interval for the population mean weight of the candies is 1.9589 , 2.0411

iii. The error bound is 0.0411

f)

i. The 98% confidence interval for the population mean weight of the candies is 1.9418 , 2.0582

iii. The error bound is 0.0582

g) The difference in the confidence intervals in part (f) and part (e) is of the level of confidence. The change is, the change in the area being calculated for the normal distribution. Therefore, the larger confidence level results in larger area and larger interval.

Hence the interval in part is larger than the one in part (e).

h) The interval in part (f) signifies that with 98% confidence that the population mean weight of the bag of candies lies between 1.942 ounce and 2.058 ounce.

5 0

3 years ago

LaVelle is making a pitcher of caffe mocha. For each ounce of chocolate syrup, she uses five ounces of coffee. How many ounces o

Nataly [62]

Answer:

40 ounces of coffee

Step-by-step explanation:

we know that

To make caffe mocha, LaVelle uses one ounce of chocolate syrup,and five ounces of coffee

so

To make 6 ounces of caffe mocha (one ounce of chocolate syrup plus 5 ounces of coffe), LaVelle uses one ounce of chocolate syrup,and five ounces of coffee

using proportion

Find out how many ounces of coffee does she need to make 48 ounces of caffe mocha

Let

x ---> ounces of coffee needed to make 48 ounces of caffe mocha

$\frac{6}{5}\ \frac{oz\ caffe\ mocha}{oz\ coffee} =\frac{48}{x}\ \frac{oz\ caffe\ mocha}{oz\ coffee}\\\\x=48(5)/6\\\\x=40\ oz\ coffee$

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3 years ago

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Law Incorporation [45]

Answer:

the answer is 17.41..

Step-by-step explanation:

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3 years ago

IF P(A)=1/3, P(B)=2/5, and P(AuB)=3/5, what is P(A^B)?

Bumek [7]

The probability of the union of two events is the sum of their probability, minus the probability of their interserction:

$P(A \cup B) = P(A) + P(B) - P(A \cap B)$

If we plug the known values into this formula, we have

$\dfrac{3}{5} = \dfrac{1}{3} + \dfrac{2}{5} - P(A\cap B)$

From which we can deduce

$P(A\cap B) = \dfrac{1}{3} + \dfrac{2}{5} - \dfrac{3}{5} = \dfrac{2}{15}$

So, the probability of $A \setminus B$ is a bit less than $P(A)$ , we have to take away all events that belong to B as well:

$P(A \setminus B) = P(A) - P(A\cap B) = \dfrac{1}{3} - \dfrac{2}{15} = \dfrac{1}{5}$

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3 years ago

Find the slope of the following equation .simplify your answer 5x+2y=-10

defon

Answer:

m=-5/2

Step-by-step explanation:

Use the slope-intercept form y=mx+b to find the slope m.

6 0

3 years ago

Read 2 more answers