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VARVARA [1.3K]
3 years ago
6

Convert 945 in/sec to miles per hour. round to one decimal​

Mathematics
1 answer:
Ulleksa [173]3 years ago
6 0

Answer:

53.7 miles per hour

divide the speed value by 17.6

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Which of the following is a counterexample to the statement “All even numbers are divisible by 4”? 2 x 2 = 4 7 is not divisible
Naily [24]
10 is not divisible by 4
5 0
3 years ago
a shipement of 14 microwave ovens contains six defective units A vending company has ordered 4 of these units and selection will
kupik [55]

Answer:

The probability that all 4 units are good is 10.66%.

Step-by-step explanation:

Given that a shipement of 14 microwave ovens contains six defective units, and a vending company has ordered 4 of these units and selection will be random, to determine what is the probability that all 4 units are good the following calculation must be performed:

14 - 6 = 8

8/14 x 8/14 x 8/14 x 8/14 = X

0.5714 x 0.5714 x 0.5714 x 0.5714 = X

0.1066 = X

0.1066 x 100 = 10.66

Therefore, the probability that all 4 units are good is 10.66%.

3 0
2 years ago
if it took 9 hours to mow 15 lawns, then at that rate, how many lawns could be mowed in 27 hours? What was the rate of lawns mow
sammy [17]

Answer:

Step-by-step explanation:

15lawns/9hours= 1.6 lawns mowed per hour

1.6lph*27hours=43.2 lawns

6 0
2 years ago
WILL GIVE BRAINLIEST THANKS AND 5 STARS FOR CORRECT ANSWER Desmond ran a lemonade stand in front of his house for the fans walki
Fiesta28 [93]

Answer:

56

Step-by-step explanation:

First, convert the 70% so that we can use it with the 80 cups. To do this, convert the percent to a decimal by moving the decimal 2 spaces to the left:

70% → 70.00 → 0.70 → 0.7

70% percent can be seen as 0.7. Multiply this by 80:

80*0.7=56

So, 70% of 80 is 56. Desmond sold 56 cups on the second Saturday.

:Done

7 0
3 years ago
Decide which chi-square test (goodness-of fit, homogeneity, or independence) would be most appropriate for the given situation.
sasho [114]

Answer:

C. Test for Goodness-of-fit.

Step-by-step explanation:

C. Test for Goodness-of-fit would be most appropriate for the given situation.

A. Test Of  Homogeneity.

The value of q is large when the sample variances differ greatly and is zero when all variances are zero . Sample variances do not differ greatly in the given question.

B. Test for Independence.

The chi square  is used to test the hypothesis about the independence of two variables each of which is classified into number of attributes. They are not classified into attributes.

C. Test for Goodness-of-fit.

The chi square test is applicable when the cell probabilities depend upon unknown parameters provided that the unknown parameters are replaced with  their estimates and provided that one degree of freedom is deducted for each parameter estimated.

4 0
2 years ago
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