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3 years ago

40 minutes as a fraction

Mathematics

2 answers:

umka21 [38]3 years ago

7 0

40/60 40 minutes is the information given and 60 is the rate of one hour

SOVA2 [1]3 years ago

6 0

4/6 or 2/3 is 40 min as a fraction

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Please help!!<br> Each pair of polygons is similar. Find the value of x.

Simora [160]

Answer in fraction form: x = 55/4

Answer in decimal form: x = 13.75

Both forms are equivalent, but they say it in a slightly different way.

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Work Shown:

AC/QY = CD/YW

(x+5)/15 = 5/4

4(x+5) = 15*5

4x+20 = 75

4x = 75-20

4x = 55

x = 55/4

x = 13.75

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Explanations:

The first step is possible because the polygons are similar. Similar figures have their corresponding sides be in the same proportion or ratio. Note how side AC has the double red angle (at point A) and a single red angle (point C). The side QY has the same idea in the same order. This is why AC corresponds to QY. The ratio AC/QY is the same as CD/YW.
In the second step, I plugged in the expressions that the diagram shows. This is the substitution property.
On step 3, I cross multiplied. The general format is A/B = C/D cross multiplies to A*D = B*C.

7 0

3 years ago

Read 2 more answers

The body weight of a healthy 3 month-old colt should be about μ = 60 kg. (Source: The Merck Veterinary Manual, a standard refere

tatuchka [14]

Answer:

The null hypothesis will be $H_o : \mu = 60 \ kg$

The alternative hypothesis will be $H_a : \mu < 60$

The alternative hypothesis will be $H_a : \mu > 60$

The alternative hypothesis will be $H_a : \mu \ne 60$

left; right; both

Step-by-step explanation:

Considering question a

The body weight of the colt is $\mu = 60 \ kg$

The null hypothesis will be $H_o : \mu = 60 \ kg$

Considering question b

Given that we want to test the claim that the average weight of a wild Nevada colt is less than 60 kg , then

The alternative hypothesis will be $H_a : \mu < 60$

Considering question c

Given that we want to test the claim that the average weight of such a wild colt is greater than 60 kg , then

The alternative hypothesis will be $H_a : \mu > 60$

Considering question d

Given that we want to test the claim that the average weight of such a wild colt is different from 60 kg , then

The alternative hypothesis will be $H_a : \mu \ne 60$

Considering question e

For question b , the corresponding p-value will be on the left because the average weight is less than 60 kg

For question c , the corresponding p-value will be on the right because the average weight is more than 60 kg

For question d , the corresponding p-value will be on both because the average weight is different from 60 kg

5 0

3 years ago

What is 30.5% as a fraction in simplest form

QveST [7]

I believe that 30.5 as a fraction is 30 and 1 half but don't take my word

7 0

3 years ago

Me podrían ayudar con esto?

xxTIMURxx [149]

Answer:

Step-by-step explanation:

This would be basically be zero/0 because anything times 0 is just nothing.

7 0

2 years ago

(05.08A)Triangle ABC is transformed to similar triangle A′B′C′ below:

Romashka-Z-Leto [24]

Answer:

1st option is correct i.e., 1 over 2

Step-by-step explanation:

Given that Triangle ABC transformed to Triangle A'B'C'.

Coordinated of Triangle ABC are A(2,6) , B(2,4) and C(4,4)

Coordinated of Triangle A'B'C' are A'(1,3), B'(1,2) and C'(2,2)

Scale factor of Dilation is the no. of times coordinates of vertices of a figure increase or decrease.i.e.,

if A(a,b) ⇒ A'(c,d)

where, a.k=c & b.k=d then k is scale factor of dilation.

Value of k is same for all vertices.

For coordinates of A & A' ,we have $2.k=1\implies k=\frac{1}{2} \:\: and \:\: 6.k=3\implies k=\frac{1}{2}$

For coordinates of B & B' ,we have $2.k=1\implies k=\frac{1}{2} \:\: and \:\: 4.k=2\implies k=\frac{1}{2}$

For coordinates of C & C' ,we have $4.k=2\implies k=\frac{1}{2} \:\: and \:\: 4.k=2\implies k=\frac{1}{2}$

Therefore, the scale factor of dilation is $\:\frac{1}{2}\:$ since in all cases $k=\frac{1}{2}$

1st option is correct i.e., 1 over 2

3 0

3 years ago