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3 years ago

There are 59 people going on a field trip.

1 answer:

6 0

The answer to this question is the school is going to have to rent 2 vans extra. The reason is because each van hols 8 people right. for the 6 vans that the school already owns, that will be 48 students in vans. When you subract that from 59, you get 11 students that still need a van, so 11 students can not fit in one extra van so the school would need two. Hope this gives you what you were looking for. God bless, and hope this helped. Also, please let me know if it was correct.

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View the screenshot and answer please. 20 points will give brainliest!

Paul [167]

Answer:

Right angle triangle goes to 15m, 8m, 17m

Step-by-step explanation:

Square root of 15^2 + 8^2 is 17

(The use of a^2 + b^2 = c^2 - Pythagoras's Theorem)

So, 10m ,11m ,15m isn't a right angle triangle

Hope this helps!

6 0

2 years ago

How to write y-2=-1/3 (x+6) in standard form and why PLZ HELP

stich3 [128]

$y-2=-\frac{1}{3}(x+6)\\y=-\frac{1}{3}(x+6)+2$

Simply move -2 to another side.

$y=-\frac{x}{3}-2+2$

Distribute -1/3 in (x+6)

$y=-\frac{x}{3}$ ---> Answer

8 0

3 years ago

Read 2 more answers

(5w-2p)(3w+p-4) anyone one?

Elis [28]

The picture shows the answer
HOPE THIS HELPS!!

3 0

2 years ago

The actual volumes of soda in quart-sized bottles can be described by a Normal model with a mean of 32.3 fluid ounces and a stan

VLD [36.1K]

Answer:The percentage of bottles expected to have a volume less than 32 or is 40.13%

Step-by-step explanation: The volumes of soda in quart soda bottles can be represented by a Nomal model with a= 32.3 oz

b=1.2 oz

Let S be the volume of randomly selected soda bottles

Y-score: S-a/b

For S=32 oz

Substitute the values of S,a and b into the equation

Y=32-32.3/1.2

Y=-0.25

Probability of bottles that have a volume less than 32 oz is

P(S<32)=P(Y<-025)= 0.40129

Percentage of bottles that have volume less than 32 oz will be

0.40127×100%=40.13%

5 0

2 years ago

Which expression is equivalent to ( 4mn / m^-2n^6)^-2a. n^6 / 16m^8b. n^10 / 16m^6c. n^10 / 8m^8d. 4m^3 / n^8?

Eva8 [605]

$\huge{(} \small{\frac{4mn}{ {m}^{ - 2} {n}^{6} }} \huge{) } \small{{}^{ {}^{ {}^{ {}^{ - 2} } } } }\\ \\ \\ \huge{(} \small{ \frac{4 \times m {}^{1} \times n {}^{1} }{ {m}^{ - 2} \times {n}^{6} }} \huge{)} \small{^{ {}^{ {}^{ {}^{ - 2} } } }} \\ \\ \\ \huge{(} \small{4 \times {m}^{1 - ( - 2)} \times {n}^{1 - 6}} \huge{)} \small{ {}^{ {}^{ {}^{ {}^{ - 2} } } } } \\ \\ \\ \huge{(} \small{ 4 \times {m}^{1 + 2} \times {n}^{ - 5} } \huge{)} \small{ { }^{ {}^{ {}^{ {}^{ - 2} } } } } \\ \\ \\ \small{ {(4)}^{ - 2} \times ({m}^{3} ) ^{ - 2} \times {(n{}^{ - 5} )}^{ - 2} } \\ \\ \\ \frac{1}{ {4}^{2} } \times {m}^{ - 6} \times {n}^{ 10} \\ \\ \\ \frac{1}{16} \times \frac{1}{ {m}^{6} } \times {n}^{10} \\ \\ \\ \frac{ {n}^{10} }{16{m}^{6} }$

$\text{ Hence, option B is correct }$

7 0

3 years ago