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3 years ago

Please answer fully

Mathematics

1 answer:

Andrews [41]3 years ago

4 0

Answer:

There is a 1/6 chance of rolling a certain number and 1/2 chance of getting a heads.

1/6 x 1/2 = 1/12 so C.

1 is the chance of getting the wanted number / 6 is all the numbers in total

When there are multiple chances you just multiply the fractions

Step-by-step explanation:

You might be interested in

Hugh bought some magizines that cost 3.95 each and some books that cost 8.95 each. He spent a total of 47.65.If Hugh bought 3 ma

HACTEHA [7]

Answer:

4 books

Step-by-step explanation:

We are told that: Hugh bought some magazines that cost 3.95 each

Hugh bought 3 magazines.

Hence, total cost of magazines = 3 magazines × 3.95

= 11.85.

Total cost of magazines = 11.85

From the question, we know that, the total amount Hugh spent = 47.65

Hence, The total cost for books =

Total amount spent - Total cost of magazines

= 47.65 - 11.85

= 35.80

Therefore, the total cost of books = 35.80

We are told in the question that the cost of books is 8.95 each or 8.95 per book.

Hence, the number of books that Hugh bought is calculated as :

35.80/8.95

= 4 books

Therefore, Hugh bought 4 books

6 0

3 years ago

1. -168 < -12a

kotegsom [21]

Answer:

See step by step explanation

Step-by-step explanation:

1. a<14

2. m </ 25

3. n >/ 7

6 0

3 years ago

Read 2 more answers

The following describes a sample. The information given includes the five number summary, the sample size, and the largest and s

lys-0071 [83]

Answer:

$L= Q_1 - 1.5*IQR = 210 -1.5*40 =150$

$U= Q_3 + 1.5*IQR = 250 +1.5*40 =310$

And if we analyze the info provided we have 500 values

Tails: 160,165,167,171,175,...,268,269,269,269,270,270

So as we can see on the tails any values is <150 or >310 so then for this case we cannot consider as outliers any of the values on the tails of the distribution.

Step-by-step explanation:

For this case we have the 5 number summary

(160,210,220,250,270)

So then we have:

minimum = 160 , Q1 = 210, Q2= Median=220, Q3 = 250, Max=270

If we find the interquartile range we got:

$IQR = Q_3 -Q_1 = 250-210 =40$

For this case we need to find the lower and upper limit with the following formulas:

$L= Q_1 - 1.5*IQR = 210 -1.5*40 =150$

$U= Q_3 + 1.5*IQR = 250 +1.5*40 =310$

And if we analyze the info provided we have 500 values

Tails: 160,165,167,171,175,...,268,269,269,269,270,270

So as we can see on the tails any values is <150 or >310 so then for this case we cannot consider as outliers any of the values on the tails of the distribution.

4 0

4 years ago

ASAP PLEASE!!!! GIVING 25 POINTSSSS

Arturiano [62]

Answer:

7b^3 + 9b^2 + 8b

Step-by-step explanation:

Combine like terms

My teacher taught me a few years ago to put like terms in the same format so they're easier to combine:

b^3 + 7b + 9b^2 + b + 6b^3

Here I have all terms to the 3rd power bolded, all terms to the 2nd power underlined, and all terms to the 1st power italicized. That makes it easier to combine those that are similar

b^3+ 6b^3 = 7b^3

9b^2 - remains the same

7b + b = 8b

Now just reorder them to where the exponents are listed from greatest to least:

7b^3 + 9b^2 + 8b

8 0

4 years ago

CAN SOMEONE PLEASE HELP ME ITS DUE TODAY

MariettaO [177]

Answer:

Step-by-step explanation:

HCF=Product of the powers of smallest prime number

HCF=5*3

HCF=15

4 0

3 years ago

Read 2 more answers

Please answer fully​

Please answer fully