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3 years ago

What is the hcf of 600 and 1050

Mathematics

1 answer:

Nata [24]3 years ago

7 0

Step-by-step explanation:

staff of 600 is 300 and the form of 1050 is 525

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a cyclist bikes a certain distance in 45 minutes , write an equation that shows the relationship between speed,s, and time,t, wh

denis23 [38]

An equation that shows the relationship between speed and time is $Speed = \frac{100}{45}$

<u>Given the following data:</u>

Distance = 100 miles.

Time = 45 minutes.

To write an equation that shows the relationship between speed and time:

<h3>What is speed?</h3>

Speed can be defined as the distance covered by a physical object per unit of time. Thus, the cyclist's speed can be measured as miles per minutes.

Mathematically, speed is given by this formula;

$Speed = \frac{distance}{time}$

Substituting the given parameters into the formula, we have;

$Speed = \frac{100}{45}$

Read more on speed here: brainly.com/question/10545161

8 0

1 year ago

7 feet =___inches can some one help me with this

ozzi

Answer:

84 in.

Step-by-step explanation:

multiply 12 by 7

there are 12 inches in a foot and 7 feet

5 0

2 years ago

Read 2 more answers

Thirty percent (30%) of the bulbs in a large box are defective. If 12 bulbs are selected randomly from the box, calculate the pr

AleksandrR [38]

The probability that exactly 6 are defective is 0.0792.

Given:

30% of the bulbs in a large box are defective.

If 12 bulbs are selected randomly from the box.

To find:

The probability that exactly 6 are defective.

Solution:

Probability of defective bulbs is:

$p=30\%$

$p=\dfrac{30}{100}$

$p=0.3$

Probability of non-defective bulbs is:

$q=1-p$

$q=1-0.3$

$q=0.7$

The probability that exactly 6 are defective is:

$P(x=r)=^nC_rp^rq^{n-r}$

$P(x=6)=^{12}C_6(0.3)^6(0.7)^{12-6}$

$P(x=6)=924(0.3)^6(0.7)^{6}$

$P(x=6)=0.0792478958$

$P(x=6)\approx 0.0792$

Therefore, the probability that exactly 6 are defective is 0.0792.

Learn more:

brainly.com/question/12917164

8 0

2 years ago

Which series of transformations will not map figure H onto itself

7nadin3 [17]

Answer:

Step-by-step explanation:

Given a square with vertices at points (2,1), (1,2), (2,3) and (3,2).

Consider option A.

1st transformation $(x+0,y-2)$ will map vertices of the square into points

$(2,1)\rightarrow (2,-1);$
$(1,2)\rightarrow (1,0);$
$(2,3)\rightarrow (2,1);$
$(3,2)\rightarrow (3,0).$

2nd transformation = reflection over y = 1 has the rule (x,2-y). So,

$(2,-1)\rightarrow (2,3);$
$(1,0)\rightarrow (1,2);$
$(2,1)\rightarrow (2,1);$
$(3,0)\rightarrow (3,2)$

These points are exactly the vertices of the initial square.

Consider option B.

1st transformation $(x+2,y-0)$ will map vertices of the square into points

$(2,1)\rightarrow (4,1);$
$(1,2)\rightarrow (3,2);$
$(2,3)\rightarrow (4,3);$
$(3,2)\rightarrow (5,2).$

2nd transformation = reflection over x = 3 has the rule (6-x,y). So,

$(4,1)\rightarrow (2,1);$
$(3,2)\rightarrow (3,2);$
$(4,3)\rightarrow (2,3);$
$(5,2)\rightarrow (1,2)$

These points are exactly the vertices of the initial square.

Consider option C.

1st transformation $(x+3,y+3)$ will map vertices of the square into points

$(2,1)\rightarrow (5,4);$
$(1,2)\rightarrow (4,5);$
$(2,3)\rightarrow (5,6);$
$(3,2)\rightarrow (6,5).$

2nd transformation = reflection over y = -x + 7 will map vertices into points

$(5,4)\rightarrow (3,2);$
$(4,5)\rightarrow (2,3);$
$(5,6)\rightarrow (1,2);$
$(6,5)\rightarrow (2,1)$

These points are exactly the vertices of the initial square.

Consider option D.

1st transformation $(x-3,y-3)$ will map vertices of the square into points

$(2,1)\rightarrow (-1,-2);$
$(1,2)\rightarrow (-2,-1);$
$(2,3)\rightarrow (-1,0);$
$(3,2)\rightarrow (0,-1).$

2nd transformation = reflection over y = -x + 2 will map vertices into points

$(-1,-2)\rightarrow (4,3);$
$(-2,-1)\rightarrow (3,4);$
$(-1,0)\rightarrow (2,3);$
$(0,-1)\rightarrow (3,2)$

These points are not the vertices of the initial square.

5 0

3 years ago

Thomas wants to invite Madeline to a party. He has an 80% chance of bumping into her at school. Otherwise, he’ll call her on the

Alexandra [31]

Answer:

84%

Step-by-step explanation:

The probability of Thomas bumping into her at school is 80%, so the probability of not bumping into her is 100% - 80% = 20%.

If he doesn't bump into her (20% chance), he will call her, and the probability of asking her in this case is 60%, so the final probability of asking her in this case is:

$P_1 = 20\% * 60\% = 12\%$

If he bumps into her (80% chance), the probability of asking her is 90%, so the final probability of asking her in this case is:

$P_2 = 80\% * 90\% = 72\%$

To find the probability of Thomas inviting Madeline to the party, we just have to sum the probabilities we found above:

$P = P_1 + P_2$

$P = 12\% + 72\% = 84\%$

5 0

3 years ago