How To work this out ?

Mathematics

1 answer:

MrMuchimi3 years ago

6 0

(a) Using the completing the square method, we need to write into the form of where , so
Expand to find the value of c
Notice that we get back the first two terms, the and the .We need to get rid of the last term of '9' as the term was not in the original form. The final form will look like
Hence,
(b)
(c) , square root both sides plus and minus of 2 Hence

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At Sanger’s auto garage, three out of every five cars brought in for service need an oil change. Of the cars that need an oil ch

Alina [70]

Answer:

The probability that a car that comes into the garage needs both an oil change and a tire rotation is $\frac{12}{35}$ .

Step-by-step explanation:

The conditional probability of an event B given that another event A has already occurred is:

$P(B|A)=\frac{P(A\cap B)}{P(A)}$

Denote the events as follows:

X = cars brought in for service need an oil change

Y = cars brought in for service need a tire rotation

The information provided is:

$P(X)=\frac{3}{5}$

$P(Y|X)=\frac{4}{7}$

Compute the value of P (X ∩ Y) as follows:

$P(Y|X)=\frac{P(X\cap Y)}{P(X)}$

$\frac{4}{7}=\frac{P(X\cap Y)}{3/5}$

$P(X\cap Y)=\frac{4}{7}\times \frac{3}{5}$

$=\frac{12}{35}$

Thus, the probability that a car that comes into the garage needs both an oil change and a tire rotation is $\frac{12}{35}$ .

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2 years ago

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What does 2miles=yards?

KiRa [710]

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A ladybug and ant move at constant speeds. The diagrams with tick marks show their positions at different times. Each tick mark

MissTica

Answer:

From the graph we can say that ant is moving faster than the ladybug

Slope of the ant's movement will be steeper than the ladybug

Hence, line "v" denotes the speed of ant and line "u" denotes the speed of ladybug.

Unit rate/slope of the graph is (distance traveled/elapsed time)

Unit rate of ant is 3/2 cm/sec and that of ladybug is 1 cm/sec.
Ant takes (12×2/3) = 8 sec and ladybug takes (12×1) = 12 sec to travel 12 cm.