Which is greater 6km, 60m, 600cm, or 6000 mm

A grocery store clerk put only cans of tomato soup and cans of chicken soup on a shelf. The ratio of the number of cans of chick

larisa [96]

Answer:

9 cans of tomato soup

Step-by-step explanation:

We are told that only cans of tomato soup and cans of chicken soup are put on the shelf.

Now, we are told that the ratio of the number of cans of chicken noodle soup to the total number of cans on the shelf was 8 to 17.

In fraction, this is 8/17

Now, this implies that there are 8 cans of chicken noodle soup while there are a total of 17 cans on the shelf.

Thus,

Since only cans of chicken noodles and cans of tomato soup are on the shelf, it means;

Cans of tomato soup = 17 - 8 = 9

8 0

3 years ago

1+3+5+7+9+11+13+15+17+19+21 without adding

alexgriva [62]

Answer:

121

Step-by-step explanation:

1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21 is a sum of first 11 odd numbers.

$\sf \boxed{\text{\bf Sum of first n odd numbers = $n^2$}}$

Here, n is the number of terms.

In this sequence, there are 11 terms.

Sum of first 11 odd numbers = 11²

= 11 *11

= 121

3 0

2 years ago

Read 2 more answers

A professor claims that his students' average score on the first exam of the semester is different than the average score on the

Marta_Voda [28]

Answer:

$t=\frac{\bar d -0}{\frac{s_d}{\sqrt{n}}}=0.704$

$p_v = 2*P(t_{7} >0.704)=0.504$

So the p value is higher the significance level given 0.1, so then we can conclude that we FAIL to reject the null hypothesis that the difference mean between after and before score is equal 0. So then we don't have enough evidence to say that the score for the second exam is different than the score for the first exam.

Step-by-step explanation:

Previous concepts

A paired t-test is used to compare two population means where you have two samples in which observations in one sample can be paired with observations in the other sample. For example if we have Before-and-after observations (This problem) we can use it.

Let put some notation

x=test value before (first exam) , y = test value after (second exam)

The system of hypothesis for this case are:

Null hypothesis: $\mu_y -\mu_x =0$

Alternative hypothesis: $\mu_y -\mu_x \neq 0$

The first step is calculate the difference $d_i=y_i-x_i$

The statistic given by :

$t=\frac{\bar d -0}{\frac{s_d}{\sqrt{n}}}=0.704$

The next step is calculate the degrees of freedom given by:

$df=n-1=8-1=7$

Now we can calculate the p value, since we have a left tailed test the p value is given by:

$p_v = 2*P(t_{7} >0.704)=0.504$

So the p value is higher the significance level given 0.1, so then we can conclude that we FAIL to reject the null hypothesis that the difference mean between after and before score is equal 0. So then we don't have enough evidence to say that the score for the second exam is different than the score for the first exam.

5 0

3 years ago

Find p and q for which the linear eqn has infinite solutions 6x-(2p-3)y-2q-3=0, 12x-( 2p-1)y-5q+1=0

astra-53 [7]

Answer:

p= 2.5

q= 7

Step-by-step explanation:

The lines should overlap to have infinite solutions, slopes should be same and y-intercepts should be same.

Equations in slope- intercept form:

6x-(2p-3)y-2q-3=0 ⇒ (2p-3)y= 6x -2q-3 ⇒ y= 6/(2p-3)x -(2q+3)/(2p-3)
12x-( 2p-1)y-5q+1=0 ⇒ (2p-1)y= 12x - 5q+1 ⇒ y=12/(2p-1)x - (5q-1)/(2p-1)

<u>Slopes equal:</u>

6/(2p-3)= 12/(2p-1)

6(2p-1)= 12(2p-3)

12p- 6= 24p - 36

12p= 30

p= 30/12

p= 2.5

<u>y-intercepts equal:</u>

(2q+3)/(2p-3)= (5q-1)/(2p-1)

(2q+3)/(2*2.5-3)= (5q-1)/(2*2.5-1)

(2q+3)/2= (5q-1)/4

4(2q+3)= 2(5q-1)

8q+12= 10q- 2

2q= 14

q= 7

3 0

4 years ago

HELP ME LLEASE WUIZ OMG WHAT ARE THE CORDS? 30 points

wariber [46]

Answer: -1/2 and 1/3

Step-by-step explanation:

3 0

3 years ago