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HELP PLEASE SOMEONE HELP MEEE ASAPPPP

diamong [38]

Answer:

5

Step-by-step explanation:

You will multiply 1/3 by 3. This equals 1. 5^1 equals 5.

7 0

3 years ago

During a test period, an experimental group of 10 vehicles using an 85 percent ethanol-gasoline mixture showed mean CO2 emission

Zinaida [17]

Answer:

-1.683

Step-by-step explanation:

Given :

Group 1 :

x1 = 667 ; n1 = 10 ; s1 = 20

Group 2 :

x2 = 679 ; n2 = 14 ; s2 = 15

The test statistic assuming equal variance :

x1 - x2 / √[Sp² * (1/n1 + 1/n2)]

sp² = [(n1 - 1)*s1² + (n2 - 1)*s2²] ÷ (n1 + n2 - 2)

Sp² = [(10 - 1)*20² + (14 - 1)*15²] = 296.59

Test statistic =

(667 - 679)/ √[296.59 * (1/10 + 1/14)]

-12 / 7.1304978

Test statistic = - 1.682

8 0

2 years ago

A dog can walk at 3miles for 10 seconds what distance will he cover

SVETLANKA909090 [29]

Answer:

Where is the rest of the equation?

Step-by-step explanation:

6 0

2 years ago

Jena's income is $1600 a month, and she plans to budget 1/3 of her income for rent and 1/8 of her income for groceries. Step 2 o

WARRIOR [948]

Answer:

Jena spends $ 733. 33 on her rent and groceries altogether per month.

Step-by-step explanation:

Jena's total income in a month= $1600

Here , expenditure on rent = (1/3) of her income

So, now one- third of her income $1600 is

= $\frac{1}{3} \times 1600 = 533.33$

or, (1/3) of her salary is $533. 33

Hence, Jena spends $ 533. 33 on her rent monthly.

Now, her expenditure on groceries = (1/8) of her income

So, now one- eighth of her income $1600 is

= $\frac{1}{8} \times 1600 = 200$

or, (1/8) of her salary is $ 200.

Jena spends $ 200. on her groceries monthly.

So, Total expenditure = Expenditure on rent + Expenditure on groceries

= $533. 33 + $200 = $733.33

Hence, Jena spends $ 733. 33 on her rent and groceries altogether per month.

7 0

2 years ago

Q‒1. [5×4 marks] a) How many three-digit numbers can be formed from the digits 0, 1, 2, 3, 4, 5, and 6? (150) b) How many three-

amid [387]

Answer:

a) 294

b) 180

c) 75

d) 174

e) 105

Step-by-step explanation:

I assume that for each problem, the first digit can't be 0.

a) There are 6 digits that can be first, 7 digits that can be second, and 7 digits that can be third.

6×7×7 = 294

b) This time, no digit can be used twice, so there are 6 digits that can be first, 6 digits that can be second, and 5 digits that can be third.

6×6×5 = 180

c) Again, each digit can only be used once, but this time, the last digit must be odd.

If only the last digit is odd, there are 3×3×3 = 27 possible numbers.

If the first and last digits are odd, there are 3×4×2 = 24 possible numbers.

If the second and last digits are odd, there are 3×3×2 = 18 possible numbers.

If all three digits are odd, there are 3×2×1 = 6 possible numbers.

The total is 27 + 24 + 18 + 6 = 75.

d) If the first digit is 3, and the second digit is 3, there are 1×1×6 = 6 possible numbers.

If the first digit is 3, and the second digit is greater than 3, there are 1×3×7 = 21 possible numbers.

If the first digit is greater than 3, there are 3×7×7 = 147 numbers.

The total is 6 + 21 + 147 = 174.

e) If the first digit is 3, and the second digit is greater than 3, then there are 1×3×5 = 15 possible numbers.

If the second digit is greater than 3, there are 3×6×5 = 90 possible numbers.

The total is 15 + 90 = 105.

6 0

2 years ago