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3 years ago

HELP!!

Mathematics

1 answer:

taurus [48]3 years ago

8 0

Lets convert

1mile=5280ft
1h=3600s

$\\ \rm\longmapsto 1mph$

$\\ \rm\longmapsto \dfrac{5280}{3600}ft/s$

$\\ \rm\longmapsto 1.46ft/s$

If rounded

$\\ \rm\longmapsto 1.5ft/s$

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The histogram below shows information about the number of hamburgers students ate in a month:

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The answer is

D. The number of students who ate 10 hamburgers or more

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3 years ago

1) A bag contains six red marbles and eight blue marbles. You randomly pick a marble and then pick a second marble without retur

Elena-2011 [213]

Answer:

You would have 4 red marbles and 8 blue marbles since you didn't return the ones you took out to the bag.

Step-by-step explanation:

6 - 2 = 4

The 8 would stay the same

There for you would have 4 red marbles and 8 blue marbles left in the bag. Totalled you have 12 marbles in the bag and 2 out of the bag. (8 + 4 = 12)

Including the marbles in and out of the bag you have 14 marbles. The 12 inside the bag + the 2 outside of the bag = 14 marbles totalled.

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3 years ago

What is the vertex form of Y = x2 - 6x + 6?

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Answer:

$\large\boxed{y=(x-3)^2-3}$

Step-by-step explanation:

The vertex form of a quadratic equation y = ax² + bx + c:

y = a(x - h)² + k

(h, k) - coordinates of a vertex

We have the equation y = x² - 6x + 6.

Convert to the vertex form:

$y=x^2-2(x)(3)+6\\\\y=\underbrace{x^2-2(x)(3)+3^2}_{(*)}-3^2+6\qquad\text{use}\ (a-b)^2=a^2-2ab+b^2\qquad(*)\\\\y=(x-3)^2-9+6\\\\y=(x-3)^2-3\to h=3,\ k=-3$

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2 years ago

a high driver stands on a cliff 150 feet above sea level. he dives into the sea and goes 15 feet below the water. what is the di

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2 years ago

Read 2 more answers

A single card is drawn from a standard 52-card deck. Let D be the event that the card drawn is a black card and let F be the eve

tester [92]

Answer:

The indicated probability of $P(D \cup F')=\frac{25}{26}$

Step-by-step explanation:

Probability of an event E to be;

P(E) = $\frac{Number of events within E}{Total number of possible outcomes}$

As per the given condition:

Total number of possible outcomes = 52 cards.

Let the event be D and F as follows;

D : Drawn card is a black card

F : Drawn card is a 10 card.

Then,

From the given condition:

P(D) = $\frac{26}{52}$ [Out of 52 cards, 26 were black] ,

P(F) = $\frac{4}{52}$ [Out of 52 cards, there are four 10 cards]

For any two events A and B we always have;

$P(A \cup B) = P(A)+P(B)-P(A \cap B)$

Now, we have to find the indicated probability:

$P(D \cup F')=P(D)+P(F')-P(D \cap F')$ ......[1]

First find the P(F');

P(F') =1-P(F) = $1-\frac{4}{52} =\frac{52-4}{52} =\frac{48}{52}$

Also, to find $P(D \cap F')$ .

We use the formula :

For any event A and B independent variable.

$P(A \cap B) =P(A) \cdot P(B)$

then;

$P(D \cap F') =P(D) \cdot P(F') = \frac{26}{52} \cdot \frac{48}{52} =\frac{24}{52}$

Now, substitute these in [1];

$P(D \cup F')=\frac{26}{52} +\frac{48}{52} -\frac{24}{52}$ = $\frac{26+48-24}{52} =\frac{50}{52} = \frac{25}{26}$

Therefore, the probability of $P(D \cup F')=\frac{25}{26}$

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3 years ago