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2 years ago

A runner completes a 400-m race in 43.9 s. In a 100-m race, he finishes in 10.4 s. In which race was his speed faster?

Mathematics

2 answers:

Sveta_85 [38]2 years ago

5 0

Answer:

100m race

Step-by-step explanation:

because 10.4 x 4 because 100m and 400m right 10.4x4=41.6 and 43.9 so obviously the 100m race is faster.

Lunna [17]2 years ago

3 0

Answer:

Faster speed is in 100-m race

Step-by-step explanation:

400/43.9 = 9.1116173121 m/s

100/10.4 = 9.6153846154 m/s

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Robert is traveling to Mexico for a family vacation. He is bringing 625 U.S. dollars with him. Robert needs to exchange his mone

Evgesh-ka [11]

Robert should get D. 8281.25 pesos in exchange.

Explanation:

Robert is bringing 625 U.S. dollars with him to Mexico. He needs to exchange the money he has with him to Mexico's currency i.e. from U.S. dollars to Mexican pesos.
The exchange rate is 13.25 Mexican pesos for each U.S. dollar. So 1 U.S. dollar = 13.25 Mexican pesos.
So to calculate how much 625 U.S. dollars is in pesos we multiply 625 with the exchange rate i.e. 13.25 pesos 625 U.S dollars = 625 × 13.25 = 8281.25 Mexican pesos.

7 0

2 years ago

Four people - Roger, Gail, Simon, and Lisa, - enter their names into a drawing. The winner receives a sweater, hat, or scarf. Th

Finger [1]

Answer:

1/3

Step-by-step explanation:

The probability of a student winning is 1/4.

The probability of a prize being selected is 1/3.

The probability of Gail winning a scarf is 1/4 × 1/3 = 1/12.

The probability of Simon winning is 1/4.

The probability of either is 1/12 + 1/4 = 1/3.

6 0

2 years ago

Which sequence of transformations produces R'S'T' from RST? On a coordinate plane, triangle R S T has points (0, 0), (negative 2

arsen [322]

Answer:

Step-by-step explanation:

Given is a triangle RST and another triangle R'S'T' tranformed from RST

Vertices of RST are (0, 0), (negative 2, 3), (negative 3, 1).

Vertices of R'S'T' are (2, 0), (0, negative 3), (negative 1, negative 1).

Comparing the corresponding vertices we find that x coordinate increased by 2 while y coordinate got the different sign.

This indicates that there is both reflection and transformation horizontally to the right by 2 units

So first shifted right by 2 units so that vertices became

(2,0) (0,3) (-1,1)

Now reflected on the line y=0 i.e. x axis

New vertices are

(2,0) (0,-3) (-1,-1)

8 0

2 years ago

A machine produces parts that are either defect free (90%), slightly defective (3%), or obviously defective (7%). Prior to shipm

AURORKA [14]

Answer:

(a) 0.0686

(b) 0.9984

Step-by-step explanation:

Given that a machine produces parts that are either defect free (90%), slightly defective (3%), or obviously defective (7%).

Let A, B, and C be the events of defect-free, slightly defective, and the defective parts produced by the machine.

So, from the given data:

P(A)=0.90, P(B)=0.03, and P(C)=0.07.

Let E be the event that the part is disregarded by the inspection machine.

As a part is incorrectly identified as defective and discarded 2% of the time that a defect free part is input.

So, $P\left(\frac{E}{A}\right)=0.02$

Now, from the conditional probability,

$P\left(\frac{E}{A}\right)=\frac{P(E\cap A)}{P(A)}$

$\Rightarrow P(E\cap A)=P\left(\frac{E}{A}\right)\times P(A)$

$\Rightarrow P(E\cap A)=0.02\times 0.90=0.018\cdots(i)$

This is the probability of disregarding the defect-free parts by inspection machine.

Similarly,

$P\left(\frac{E}{A}\right)=0.40$

and $\Rightarrow P(E\cap B)=0.40\times 0.03=0.012\cdots(ii)$

This is the probability of disregarding the partially defective parts by inspection machine.

$P\left(\frac{E}{A}\right)=0.98$

and $\Rightarrow P(E\cap C)=0.98\times 0.07=0.0686\cdots(iii)$

This is the probability of disregarding the defective parts by inspection machine.

(a) The total probability that a part is marked as defective and discarded by the automatic inspection machine

$=P(E\cap C)$

$= 0.0686$ [from equation (iii)]

(b) The total probability that the parts produced get disregarded by the inspection machine,

$P(E)=P(E\cap A)+P(E\cap B)+P(E\cap C)$

$\Rightarrow P(E)=0.018+0.012+0.0686$

$\Rightarrow P(E)=0.0986$

So, the total probability that the part produced get shipped

$=1-P(E)=1-0.0986=0.9014$

The probability that the part is good (either defect free or slightly defective)

$=\left(P(A)-P(E\cap A)\right)+\left(P(B)-P(E\cap B)\right)$

$=(0.9-0.018)+(0.03-0.012)$

$=0.9$

So, the probability that a part is 'good' (either defect free or slightly defective) given that it makes it through the inspection machine and gets shipped

$=\frac{\text{Probabilily that shipped part is 'good'}}{\text{Probability of total shipped parts}}$

$=\frac{0.9}{0.9014}$

$=0.9984$

(c) The probability that the 'bad' (defective} parts get shipped

=1- the probability that the 'good' parts get shipped

=1-0.9984

=0.0016

5 0

2 years ago

X=a-2by(2) make y the subject

sladkih [1.3K]

Answer:

y = √{(a - x)/2b}

Step-by-step explanation:

x=a-2by²

2by² = a - x

divide through by 2b

y² = (a - x)/2b

y = √{(a - x)/2b}

3 0

2 years ago