Solve the following multiplication problems. a. 9kg. 3hg. 1cg× 9 b. 11dg. 4cg. × 12 c. 10g. 5 dg. ×6. 5g 3dg. 4cg. × 8

Nutka1998 [239]

Answer:

$\frac{35}{48}$

Step-by-step explanation:

Probability: $Probability=\frac{favourable\ outcome}{total\ outcome}$

Probability of NOT 6:

$Total\ outcomes=\{1,2,3,4,5,6\}\\\\Number\ of\ total\ outcomes=6\\\\Favourable\ Outcomes= NOT\ 6=\{1,2,3,4,5\}\\\\Number\ of\ favourable\ outcomes=5\\\\P(NOT\ 6)=\frac{5}{6}$

Probability of NOT H:

$Total\ outcomes=\{A,B,C,D,E,F,G,H\}\\\\Number\ of\ total\ outcomes=8\\\\Favourable\ Outcomes= NOT\ H=\{A,B,C,D,E,F,G\}\\\\Number\ of\ favourable\ outcomes=7\\\\P(NOT\ H)=\frac{7}{8}$

Probability of NOT 6 and NOT H:

$Both\ events\ are\ independent\\\\P(NOT\ 6\ and\ NOT\ H)=P(NOT\ 6)\times P(NOT\ H)\\\\P(NOT\ 6\ and\ NOT\ H)=\frac{5}{6}\times \frac{7}{8}\\\\P(NOT\ 6\ and\ NOT\ H)=\frac{35}{48}$

6 0

2 years ago

Gina has a jar of marbles that contains 7 blue marbles, 5 red marbles, and 8 green marbles. She randomly selects a marble, recor

BigorU [14]

The probability that Gina randomly selected two red marbles is 1/19

Explanation:

Total number of marbles = 7 + 5 + 8

= 20

The probability of getting two red marbles in the fraction form is given as:

P(first red marble) = number of red marbles / total number of marbles

P(first red marble) = $\frac{5}{20} = \frac{1}{4}$

P(second red marble) = number of red marbles after 1 white marble is removed / total number of marbles after 1 red marble is removed.

P(first red marble) = $\frac{4}{19}$

P(two red marbles) = P(first) X P(second)

= $\frac{1}{4} X \frac{4}{19}$

= $\frac{1}{19}$

Therefore, the probability that Gina randomly selected two red marbles is 1/19

6 0

3 years ago

An education firm measures the high school dropout rate as the percentage of 16 to 24 year olds who are not enrolled in school a

posledela

Answer:

The value of the test statistic $z = -0.6606$

Step-by-step explanation:

From the question we are told that

The high dropout rate is $\mu = 6.1$ % $= 0.061$

The sample size is $n = 1000$

The number of dropouts $x = 56$

The probability of having a dropout in 1000 people $\= x = \frac{56}{1000} = 0.056$

Now setting up Test Hypothesis

Null $H_o : p = 0.061$

Alternative $H_a : p < 0.061$

The Test statistics is mathematically represented as

$z = \frac{\= x - p}{\sqrt{\frac{p(1 -p)}{n} } }$

substituting values

$z = \frac{0.056 - 0.061}{\sqrt{\frac{0.061(1 -0.061)}{1000} } }$

$z = -0.6606$

6 0

3 years ago

Given: AC bisects angle BCD , Angle ABC is congruent to angle ADC prove AB is congruent to AD. Help immediately

weeeeeb [17]

AC bisects ∠BAD, => ∠BAC=∠CAD ..... (1)

thus in ΔABC and ΔADC, ∠ABC=∠ADC (given),

∠BAC=∠CAD [from (1)],

AC (opposite side side of ∠ABC) = AC (opposite side side of ∠ADC), the common side between ΔABC and ΔADC

Hence, by AAS axiom, ΔABC ≅ ΔADC,

Therefore, BC (opposite side side of ∠BAC) = DC (opposite side side of ∠CAD), since (1)

Hence, BC=DC proved.

7 0

3 years ago

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The axis of symmetry for the graph of the function is f(x) = x2 + bx + 10 is x = 6. What is the value of b?

joja [24]

Answer:

$b=-12$

Step-by-step explanation:

we have

$f(x)=x^{2}+bx+10$

we know that

The equation of a vertical parabola into vertex form is equal to

$y=a(x-h)^{2} +k$

where

(h,k) is the vertex of the parabola

and the axis of symmetry is equal to $x=h$

In this problem we have the axis of symmetry $x=6$

so

the x-coordinate of the vertex is equal to $6$

therefore

For $x=6+1=7$ -----> one unit to the right of the vertex

Find the value of $f(7)$

$f(7)=7^{2}+b(7)+10$

$f(7)=59+7b$

For $x=6-1=5$ -----> one unit to the left of the vertex

Find the value of $f(5)$

$f(5)=5^{2}+b(5)+10$

$f(5)=35+5b$

Remember that

$f(5)=f(7)$ ------> the x-coordinates are at the same distance from the axis of symmetry

so

$59+7b=35+5b$ ------> solve for b

$7b-5b=35-59$

$2b=-24$

$b=-12$

3 0

3 years ago

Read 2 more answers