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Svetllana [295]

2 years ago

15

clayton uses 34 of a bottle of hot sauce for every batch of salsa he makes. yesterday, he used 214 bottles of hot sauce. how man

y batches of salsa did he make?

1 answer:

Aleks [24]2 years ago

8 0

If he used 34 bottles for every batch just divide 214 bottles he used by 34 it takes for a batch
he made 6.3 batches

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Jose surveyed the length of TV commercials (in seconds). Find, to 3 decimal places, the experimental probability that a randomly

Lynna [10]

Answer:

≈0.487

Step-by-step explanation:

I think this is your full question right?

Jose surveyed the length of TV commercials (in seconds). Find, to 3 decimal places, the experimental probability that a randomly chosen TV commercial will last:

Length Frequency

0-19 17

20-39 38

40-59 19 20 to 39 seconds.

60+ 20 to 4

Here is my anwer:

experimental probability = relative frequency = $\frac{number of oserved occuerencies}{total number of oserved occuerencies}$

Total frequencies = 17 + 38 +18 + 4 = 78

P(20 to 39s) = 38/78 ≈0.487

6 0

2 years ago

Please help! Question above

Gala2k [10]

Answer:

AAS

Step-by-step explanation:

The first A:

∠ABC ≅ ∠DCB

The S:

CB ≅ BC

The second A:

∠A ≅ ∠D

All of them combined to prove ΔABC ≅ ΔDCB through ASA.

8 0

2 years ago

What is 8% of 8393.....

svlad2 [7]

The answer to this question is:

671.44

7 0

2 years ago

Read 2 more answers

What is the answer????

xz_007 [3.2K]

Answer:

36/48 or 6/8 or 3/4 or .75

Step-by-step explanation:

6^2/ [(2+4)*2^3]

6^2/ [ ( 6)*2^3]

36/[(6)*8]

36/48

8 0

3 years ago

in a standard casino dice game the roller wins on the first roll if he rolls a sum of 7 or 11. what is the probability of winnin

Alborosie

<u>Answer-</u>

<em>The probability of winning on the first roll is </em><em>0.22</em>

<u>Solution-</u>

As in the game of casino, two dice are rolled simultaneously.

So the sample space would be,

$|S|=6^2=36$

Let E be the event such that the sum of two numbers are 7, so

E = {(1,6), (2,5), (3,4), (4,3), (5,2), (6,1)}

$|E|=56$

$\therefore P(E)=\dfrac{|E|}{|S|}=\dfrac{6}{36}$

Let F be the event such that the sum of two numbers are 11, so

F = {(6,5), (5,6)}

$|F|=2$

$\therefore P(F)=\dfrac{|F|}{|S|}=\dfrac{2}{36}$

Now,

$P(\text{sum is 7 or 11)}=P(E\ \cup\ F)=P(E)+P(F)=\dfrac{6}{36}+\dfrac{2}{36}=\dfrac{8}{36}=\dfrac{2}{9}=0.22$

8 0

3 years ago