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2 years ago

8

A clydesdale drinks about 120 gallons of water every 4 days.At this rate,about how many gallons of water does a clydesdale drink

in 28 days

1 answer:

il63 [147K]2 years ago

3 0

A clydes dale can drink up to 840 gallons of water in 28 days

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Find the number that belongs in the green box.<br><br>Round your answer to the nearest tenth

Umnica [9.8K]

Answer:

134.5 degree

Step-by-step explanation:

first find the remaining side of a triangle

BY using cosine rule

C^2=A^2+B^2-2ABcosx

apply square root both sides

√c^2 =√[4^2+10^2-2(4)(10)cos29]

C=√(116-69.96958)

C=√46.03

C=6.8

Now find angle using cosine rule

C^2=A^2+B^2-2ABcosx

10^2=4^2+6.78^2-2(4)(6.78)cosx

100=16+45.9684-54.24cosx

100=61.9684-54.24cosx

100-61.9684=-54.24cosx

(38.0316)/-54.24=(-54.4cosx)/-54.24

-0.70117=cosx

X= cos inverse of -0.70117

x=134.5 degree

5 0

2 years ago

Read 2 more answers

A store has a sale for 15% off. You want a jacket that was originally $45. How much

Travka [436]

Answer:

es 51.75

Step-by-step explanation:

por que al sumar 45 + 15 porciento da 51.75

4 0

2 years ago

What is the input variable,x, for this situation?

Radda [10]

Answer:

input

Step-by-step explanation

Because it is a causal relationship between x and y

4 0

3 years ago

A scientist inoculates mice, one at a time, with a disease germ until he finds 3 that have contracted the disease. If the prob

UkoKoshka [18]

Answer:

The probability that 8 mice are required is 0.2428.

Step-by-step explanation:

Given : A scientist inoculates mice, one at a time, with a disease germ until he finds 3 that have contracted the disease. If the probability of contracting the disease is two sevenths.

To find : What is the probability that 8 mice are required? The probability that that 8 mice are required is nothing ?

Solution :

Applying binomial distribution,

$P(X=r)=^nC_r p^rq^{n-r}$

Where, p is the probability of success $p=\frac{2}{7}$

q is the probability of failure q=1-p, $q=1-\frac{2}{7}=\frac{5}{7}$

n is total number of trials n=8

r=3

Substitute the values,

$P(X=3)=^8C_3 (\frac{2}{7})^3 (\frac{5}{7})^{8-3}$

$P(X=3)=\frac{8!}{3!5!}\times \frac{8}{343}\times (\frac{5}{7})^{5}$

$P(X=3)=\frac{8\times 7\times 6}{3\times 2\times 1}\times \frac{8}{343}\times \frac{3125}{16807}$

$P(X=3)=0.2428$

Therefore, the probability that 8 mice are required is 0.2428.

5 0

3 years ago

What is the total value of the digit 5 in the number 57,634

melomori [17]

Hey there!

$\bold{57,634}$

$\bold{5:}$ $\bold{ten}$ $\bold{thousands}$
$\bold{7:}$ $\bold{thousands}$
$\bold{6:}$ $\bold{hundreds}$
$\bold{3:}$ $\bold{tens}$
$\bold{4:}$ $\bold{ones}$
$\boxed{\boxed{\bold{Answer:ten-thousand(s)}}}$

Good luck on your assignment and enjoy your day!

~ $\bold{LoveYourselfFirst:)}$

3 0

3 years ago

Read 2 more answers