All categories

3 years ago

Hello can you guys help me with these math problem please!!

2 answers:

8 0

To get the total number of sandwiches, multiply the amount of sandwiches * the number of sandwiches/tray *number of types of sandwiches

5*6*5=150 sandwiches total

divided by 10 guests=15 sandwiches
yes, Cassandra is correct

sp2606 [1]3 years ago

7 0

The total number of sandwiches Cassandra ordered is 150, however Cassandra is wrong, each guest will only get 3 sandwiches.

You might be interested in

What does the 50th term of this arithmetic sequence: 29,32,35,38

frozen [14]

Answer:

176

Step-by-step explanation:

First term (a) = 29

Common difference (d) = 32 - 29 = 3

For calculating the 50th term , n = 50

Now

$t_{50} = a + ( n - 1) d\\t_{50} = 29 + ( 50 - 1 ) *3\\ t_{50} = 29 + 147\\t_{50} = 176$

hope it will help :)

3 0

3 years ago

Find the value of given expression<br><br><img src="https://tex.z-dn.net/?f=%20%5Csqrt%7B24%20%5Ctimes%205%20%2B%201%7D%20" id="

Orlov [11]

Answer:

your answer will be 11

Step-by-step explanation:

Thanks again!!

4 0

3 years ago

Read 2 more answers

Solve for x<br><br> 1/3x=7/6 + 1/9

zimovet [89]

Answer:

x=26/3 because you need to comine like terms as usual

3 0

3 years ago

1) Find the 95 % confidence interval for the mean if σ = 11 , using this sample:

melamori03 [73]

Answer:

The 95 % confidence interval for the mean is between 28.09 and 35.97.

Step-by-step explanation:

Mean of the sample:

30 values.

So we sum all the values and divide by 30.

The sum of all the values(43 + 52 + 18 + ... + 20 + 41) is 961.

961/30 = 32.03

Confidence interval:

We have that to find our $\alpha$ level, that is the subtraction of 1 by the confidence interval divided by 2. So:

$\alpha = \frac{1-0.95}{2} = 0.025$

Now, we have to find z in the Ztable as such z has a pvalue of $1-\alpha$ .

So it is z with a pvalue of $1-0.025 = 0.975$ , so $z = 1.96$

Now, find the margin of error M as such

$M = z*\frac{\sigma}{\sqrt{n}}$

In which $\sigma$ is the standard deviation of the population and n is the size of the sample.

$M = 1.96\frac{11}{30} = 3.94$

The lower end of the interval is the sample mean subtracted by M. So it is 32.03 - 3.94 = 28.09

The upper end of the interval is the sample mean added to M. So it is 32.03 + 3.94 = 35.97

The 95 % confidence interval for the mean is between 28.09 and 35.97.

8 0

3 years ago

PLEASE! HELP ASAP! Which does not explain the effect of the author's use of dialogue?

Zarrin [17]

Answer:

do you need all work shown for this problem

4 0

3 years ago