Tamir bought 2 1/2 pounds of fish at $5.50 per pound, and two bananas at the price shown

Find the slope of the line passing through the points (-7 ,-7) and (-3,6)

ycow [4]

Answer:

14/4 or 7/2

Step-by-step explanation:

6 - (-7)

---------

-3 - (-7)

When you subtract a - its really just adding

6 0

2 years ago

58.15 If f(x) = 5x + 40, what is f(x) when x = -5?

Ket [755]

Answer:

15

Step-by-step explanation:

if you put x=-15

f(x) = 5*-5 +40

=-25+40

=15

7 0

2 years ago

What stays the same no matter what location your in

nikitadnepr [17]

The mass of an object

5 0

2 years ago

A public bus company official claims that the mean waiting time for bus number 14 during peak hours is less than 10 minutes. Kar

ch4aika [34]

Answer:

We conclude that the mean waiting time is less than 10 minutes.

Step-by-step explanation:

We are given that a public bus company official claims that the mean waiting time for bus number 14 during peak hours is less than 10 minutes.

Karen took bus number 14 during peak hours on 18 different occasions. Her mean waiting time was 7.8 minutes with a standard deviation of 2.5 minutes.

Let $\mu$ = mean waiting time for bus number 14.

So, Null Hypothesis, $H_0$ : $\mu \geq$ 10 minutes {means that the mean waiting time is more than or equal to 10 minutes}

Alternate Hypothesis, $H_A$ : $\mu$ < 10 minutes {means that the mean waiting time is less than 10 minutes}

The test statistics that would be used here One-sample t test statistics as we don't know about the population standard deviation;

T.S. = $\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }$ ~ $t_n_-_1$

where, $\bar X$ = sample mean waiting time = 7.8 minutes

s = sample standard deviation = 2.5 minutes

n = sample of different occasions = 18

So, test statistics = $\frac{7.8-10}{\frac{2.5}{\sqrt{18} } }$ ~ $t_1_7$

= -3.734

The value of t test statistics is -3.734.

Now, at 0.01 significance level the t table gives critical value of -2.567 for left-tailed test.

Since our test statistic is less than the critical value of t as -3.734 < -2.567, so we have sufficient evidence to reject our null hypothesis as it will fall in the rejection region due to which we reject our null hypothesis.

Therefore, we conclude that the mean waiting time is less than 10 minutes.

5 0

3 years ago

The following function represents the profit P(n), in dollars, that a concert promoter makes by selling tickets for n dollars ea

Gre4nikov [31]

A) zeroes

P(n) = -250 n^2 + 2500n - 5250

Extract common factor:

P(n)= -250 (n^2 - 10n + 21)

Factor (find two numbers that sum -10 and its product is 21)

P(n) = -250(n - 3)(n - 7)

Zeroes ==> n - 3 = 0 or n -7 = 0
Then n = 3 and n = 7 are the zeros.

They rerpesent that if the promoter sells tickets at 3 or 7 dollars the profit is zero.

B) Maximum profit

Completion of squares

n^2 - 10n + 21 = n^2 - 10n + 25 - 4 = (n^2 - 10n+ 25) - 4 = (n - 5)^2 - 4

P(n) = - 250[(n-5)^2 -4] = -250(n-5)^2 + 1000

Maximum ==> - 250 (n - 5)^2 = 0 ==> n = 5 and P(5) = 1000

Maximum profit =1000 at n = 5

C) Axis of symmetry

Vertex = (h,k) when the equation is in the form A(n-h)^2 + k

Comparing A(n-h)^2 + k with - 250(n - 5)^2 + 1000

Vertex = (5, 1000) and the symmetry axis is n = 5.

8 0

3 years ago