If y=-4+11 and 3x+y=9 what is the value of y

umka21 [38]

<span>X^2= 1/100
x = 1/10

Because 1/10 * 1/10 = 1/100

</span>

5 0

3 years ago

The mayor is interested in finding a 98% confidence interval for the mean number of pounds of trash per person per week that is

saul85 [17]

Solution :

Given :

Sample mean, $\overline X = 34.2$

Sample size, n = 129

Sample standard deviation, s = 8.2

a. Since the population standard deviation is unknown, therefore, we use the t-distribution.

b. Now for 95% confidence level,

α = 0.05, α/2 = 0.025

From the t tables, T.INV.2T(α, degree of freedom), we find the t value as

t =T.INV.2T(0.05, 128) = 2.34

Taking the positive value of t, we get

Confidence interval is ,

$\overline X \pm t \times \frac{s}{\sqrt n}$

$34.2 \pm 2.34 \times \frac{8.2}{\sqrt {129}}$

(32.52, 35.8)

95% confidence interval is (32.52, 35.8)

So with $95 \%$ confidence of the population of the mean number of the pounds per person per week is between 32.52 pounds and 35.8 pounds.

c. About $95 \%$ of confidence intervals which contains the true population of mean number of the pounds of the trash that is generated per person per week and about $5 \%$ that doe not contain the true population of mean number of the pounds of trashes generated by per person per week.

4 0

3 years ago

The perimeter of a rectangle is 150 cm. the length is 4 cm greater than the width find the width and length

Fudgin [204]

Let's call the width of our rectangle $w$ and the length $l$ . We can say $l = w + 4$ , since the length is equal to 4 cm greater than the width.

Also remember that the perimeter of a rectangle is the sum of two times the width and two times the length, or $P = 2l + 2w$ . To solve this problem, we can substitute in the information we know, as shown below:

$150 = 2(w + 4) + 2w$

$150 = 4w + 8$

$142 = 4w$

$w = \frac{71}{2}$

Now, we can substitute in the width we found into the formula for length, which is $l = w + 4$ :

$l = \frac{71}{2} + 4$

$l = \frac{79}{2}$

The width of our rectangle is $\boxed{\frac{71}{2} \,\, \textrm{cm}}$ cm and the length of our rectangle is $\boxed{\frac{79}{2} \,\, \textrm{cm}}$

8 0

3 years ago

Or the past 100 years, the average high temperature on october 1 is 78° with a standard deviation of 5°. what is the probability

valina [46]

Answer:

The probability is 0.683

Step-by-step explanation:

To calculate this, we shall be needing to calculate the z-scores of both temperatures

mathematically;

z-score = (x-mean)/SD

From the question mean = 78 and SD = 5

For 73

z-score = (73-78)/5 = -5/5 = -1

For 83

z-score = (83-78)/5 = 5/5 = 1

So the probability we want to calculate is within the following range of z-scores;

P(-1 <z <1 )

Mathematically, this is same as ;

P(z<1) - P(z<-1)

Using the normal distribution table;

P(-1<z<1) = 0.68269 which is approximately 0.683

3 0

3 years ago

Consider the two functions described below. Function 1 describes the cost, y, to park in a parking lot for x hours. There is a f

Len [333]

Statement B explains correctly that the parking garage charges a higher rate per hour than the parking lot. Statements B and D are the same.

Step-by-step explanation:

Step 1:

For the parking lot, there is a flat fee of $7.00 and then a charge of $1.00 every hour.

So if a vehicle was parked for 1 hour charges are 7 + 1 (1) = $8,

If parked for two hours charges are 7 + 1 (2) = $9.

Step 2;

To understand the pricing of the parking garage, we must plot the points of the graph.

The points on the graph are (0, 4), (1, 6), (2, 8), (3, 10), (4, 12), (5, 14)...

So for the parking garage, there is a flat fee of $4 and then a charge of $2.00 every hour.

So if a vehicle was parked for 1 hour charges are 4 + 2 (1) = $6,

If parked for two hours charges are 4 + 2 (2) = $8.

Step 3:

Comparing both the parking lot and the parking garage, we get that the flat fee for the parking lot is higher while the parking charges for an hour are higher for the parking garage which is statement B.

Statement B and D are the same.

7 0

3 years ago