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4 years ago

Need help fast plz Subtract. 5/9−(−3 1/3)

Mathematics

2 answers:

Ahat [919]4 years ago

7 0

Answer would be 35/9

ICE Princess25 [194]4 years ago

3 0

Answer you seek is: 3 <span>8/<span>9

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Help me please thanks:)))

storchak [24]

The combined weight of the pumpkins would be
74 and 15/16 pounds

7 0

3 years ago

Using Cramer’s Rule, what is the value of x in the system of linear equations below?

cluponka [151]

Answer:

x = 8 is the answer

Step-by-step explanation:

In order to find the solution of Equations using Cramer's rule

we shall find matrices

D= $\left[\begin{array}{ccc}5&-3\\1&-2\end{array}\right]$ =

$D_{1} = \left[\begin{array}{ccc}4&-3\\-16&-2\end{array}\right]$

l D l = Determinant formed by coefficient of variables

Here l D l = 5X(-2) - 1X(-3)= -10+3 = -7

l D1 l = Determinant formed by replacing first column In D by numbers on the right side of the equations

Here l D1 l = 4X(-2)- (-16)X(-3)= -8-48 = -56

here x is given by $\frac{l D1 l}{l D l}$

x = $\frac{-56}{-7}$

x = 8

3 0

3 years ago

What is the solution to the system of eqautions? y=2/3x+3, x=-2

morpeh [17]

Answer:
(-2, 5/3)
Step-by-step explanation:
Step 1: Write out systems of equations
y = 2/3x + 3
x = -2
Step 2: Substitution
y = 2/3(-2) + 3
y = -4/3 + 3
y = 5/3
Step 3: Write solution
(-2, 5/3)

8 0

3 years ago

I don't understand math @ all please help

chubhunter [2.5K]

More than likely it is c, because I looked over all the answers and c was the best one for the problem

7 0

4 years ago

During the 2015-16 NBA season, J.J. Redick of the Los Angeles Clippers had a free throw shooting percentage of 0.901 . Assume th

ser-zykov [4K]

Answer: 0.5898

Step-by-step explanation:

Given : J.J. Redick of the Los Angeles Clippers had a free throw shooting percentage of 0.901 .

We assume that,

The probability that .J. Redick makes any given free throw =0.901 (1)

Free throws are independent.

So it is a binomial distribution .

Using binomial probability formula, the probability of getting success in x trials :

$P(X=x)^nC_xp^x(1-p)^{n-x}$

, where n= total trials

p= probability of getting in each trial.

Let x be binomial variable that represents the number of a=makes.

n= 14

p= 0.901 (from (1))

The probability that he makes at least 13 of them will be :-

$P(x\geq13)=P(x=13)+P(x=14)$

$=^{14}C_{13}(0.901)^{13}(1-0.901)^1+^{14}C_{14}(0.901)^{14}(1-0.901)^0\\\\=(14)(0.901)^{13}(0.099)+(1)(0.901)^{14}\ \ [\because\ ^nC_n=1\ \&\ ^nC_{n-1}=n ]\\\\\approx0.3574+0.2324=0.5898$

∴ The required probability = 0.5898

5 0

3 years ago