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geniusboy [140]

3 years ago

15

X + 6.2 10.9 can you solve it plz i need can you explain ste by step and not to long plz

2 answers:

kari74 [83]3 years ago

8 0

Answer:

so u it is

Step-by-step explanation:

Masteriza [31]3 years ago

8 0

I believe that its X=4.7

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32 is 40% of what number?<br><br><br><br> Enter your answer in the box.

Rufina [12.5K]

34 / .40 = 85

I hope this helps :)

6 0

3 years ago

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Which of the following gives a product of x^3 - 3x ?

Vilka [71]

Answer:

Option a

Step-by-step explanation:

Lets verify

RHS

$\\ \rm\longmapsto x(x^2-3)$

Use distributive law
a(b-c)=ab-bc

$\\ \rm\longmapsto x(x^2)-3x$

$\\ \rm\longmapsto x^3-3x$

$\\ \rm\longmapsto LHS$

Hence verifed

7 0

3 years ago

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If X = 2 centimeters, Y = 5 centimeters, and Z = 5 centimeters, what is the area of the object?

Elena L [17]

Answer: $35 cm^2$

Step-by-step explanation:

<em>Area of trapezium = </em>

2X = 2 + 2 = 4 cm

2Z = 5 + 5 = 10 cm

Y = 5 cm

$\dfrac{ \left( 4+10 \right) }{ 2 } \times 5$

|

|--------> 35 $cm^2$

6 0

1 year ago

Required information Skip to question A die (six faces) has the number 1 painted on two of its faces, the number 2 painted on th

grigory [225]

Answer:

The change to the face 3 affects the value of P(Odd Number)

Step-by-step explanation:

Analysing the question one statement at a time.

Before the face with 3 is loaded to be twice likely to come up.

The sample space is:

$S = \{1,1,2,2,2,3\}$

And the probability of each is:

$P(1) = \frac{n(1)}{n(s)}$

$P(1) = \frac{2}{6}$

$P(1) = \frac{1}{3}$

$P(2) = \frac{n(2)}{n(s)}$

$P(2) = \frac{3}{6}$

$P(2) = \frac{1}{2}$

$P(3) = \frac{n(3)}{n(s)}$

$P(3) = \frac{1}{6}$

P(Odd Number) is then calculated as:

$P(Odd\ Number) = P(1) + P(3)$

$P(Odd\ Number) = \frac{1}{3} + \frac{1}{6}$

Take LCM

$P(Odd\ Number) = \frac{2+1}{6}$

$P(Odd\ Number) = \frac{3}{6}$

$P(Odd\ Number) = \frac{1}{2}$

After the face with 3 is loaded to be twice likely to come up.

The sample space becomes:

$S = \{1,1,2,2,2,3,3\}$

The probability of each is:

$P(1) = \frac{n(1)}{n(s)}$

$P(1) = \frac{2}{7}$

$P(2) = \frac{n(2)}{n(s)}$

$P(2) = \frac{3}{7}$

$P(3) = \frac{n(3)}{n(s)}$

$P(3) = \frac{1}{7}$

$P(Odd\ Number) = P(1) + P(3)$

$P(Odd\ Number) = \frac{2}{7} + \frac{1}{7}$

Take LCM

$P(Odd\ Number) = \frac{2+1}{7}$

$P(Odd\ Number) = \frac{3}{7}$

Comparing P(Odd Number) before and after

$P(Odd\ Number) = \frac{1}{2}$ --- Before

$P(Odd\ Number) = \frac{3}{7}$ --- After

<em>We can conclude that the change to the face 3 affects the value of P(Odd Number)</em>

7 0

3 years ago

Carlos has 38 apples.He puts 8 into each bag.If he fills many bags as possible ,how many apples are left?

leva [86]

Well that would be 8 apples in 4 bags so there’s 6

answer: 6

7 0

3 years ago