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3 years ago

Please help with #3 this is the last one to complete my HW!!

Mathematics

2 answers:

Alex_Xolod [135]3 years ago

8 0

10/5+10-9 * 11
=2+10-99
=12-99
=87

Zepler [3.9K]3 years ago

6 0

10 divided by 5 is 2 +10=12,subtract 9 is 3 x 11=33.

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Whilst shopping, the probability that Caroline buys fruit is 0.7.

Lilit [14]

Answer:

Probability that Caroline buys fruit, a CD or both is 0.76.

Step-by-step explanation:

Let event A = Caroline buys fruit, event B = Caroline buys CD, Ac and Bc are complementary events.

Events AB, ABc, AcB and AcBc are jointly exhaustive and disjoint, hence P(AB) + P(ABc) + P(AcB) +P(AcBc) =1.

Events A and B independent, hence Ac and Bc independent too and probability P(AcBc) = P(Ac)*P(Bc) = (1 - P(A))(1-P(B)) = 0.6*0.4 = 0.24.

Required probability P(AB + ABc + AcB ) = P(AB) + P(ABc) + P(AcB) = 1- P(AcBc) = 1 - 0.24 = 0.76.

3 0

3 years ago

Read 2 more answers

What types of solutions does 6x^2 - 20x + 1 have?

elena55 [62]

Answer:

2 real solutions

Step-by-step explanation:

We can use the determinant, which says that for a quadratic of the form ax² + bx + c, we can determine what kind of solutions it has by looking at the determinant of the form:

b² - 4ac

If b² - 4ac > 0, then there are 2 real solutions. If b² - 4ac = 0, then there is 1 real solution. If b² - 4ac < 0, then there are 2 imaginary solutions.

Here, a = 6, b = -20, and c = 1. So, plug these into the determinant formula:

b² - 4ac

(-20)² - 4 * 6 * 1 = 400 - 24 = 376

Since 376 is clearly greater than 0, we know this quadratic has 2 real solutions.

~ an aesthetics lover

3 0

2 years ago

Two competitive neighbours build rectangular pools that cover the same area but are different shapes. Pool A has a width of (x +

GenaCL600 [577]

Answer:

a)Dimensions of pool A are length = 6.667m and width = 3.667 m and dimension of pool B are length = 7.333m and width = 3.333m.

b) Area of pool A is equal to area of pool B equal to 24.44 meters.

Solution:

Let’s first calculate area of pool A .

Given that width of the pool A = (x+3)

Length of the pool A is 3 meter longer than its width.

So length of pool A = (x+3) + 3 =(x + 6)

Area of rectangle = length x width

So area of pool A =(x+6) (x+3) ------(1)

Let’s calculate area of pool B

Given that length of pool B is double of width of pool A.

So length of pool B = 2(x+3) =(2x + 6) m

Width of pool B is 4 meter shorter than its length,

So width of pool B = (2x +6 ) – 4 = 2x + 2

Area of rectangle = length x width

So area of pool B =(2x+6)(2x+2) ------(2)

Since area of pool A is equal to area of pool B, so from equation (1) and (2)

(x+6) (x+3) =(2x+6) (2x+2)

On solving above equation for x

(x+6) (x+3) =2(x+3) (2x+2)

x+6 = 4x + 4

x-4x = 4 – 6

x = $\frac{2}{3}$

Dimension of pool A

Length = x+6 = ( $\frac{2}{3}$ ) +6 = 6.667m

Width = x +3 = ( $\frac{2}{3}$ ) +3 = 3.667m

Dimension of pool B

Length = 2x +6 = 2( $\frac{2}{3}$ ) + 6 = $\frac{22}{3}$ = 7.333m

Width = 2x + 2 = 2( $\frac{2}{3}$ ) + 2 = $\frac{10}{3}$ = 3.333m

Verifying the area:

Area of pool A = ( $\frac{20}{3}$ ) x ( $\frac{11}{3}$ ) = $\frac{220}{9}$ = 24.44 meter

Area of pool B = ( $\frac{22}{3}$ ) x ( $\frac{10}{3}$ ) = $\frac{220}{9}$ = 24.44 meter

Summarizing the results:

(a)Dimensions of pool A are length = 6.667m and width = 3.667 m and dimension of pool B are length = 7.333m and width = 3.333m.

(b)Area of pool A is equal to Area of pool B equal to 24.44 meters.

5 0

3 years ago

A box of facial tissues is 22 cm by 10.5 cm by 8 cm how much cardboard is on the outside surface?

beks73 [17]

Dude y you even as its just the surface area you just use the formula

4 0

3 years ago

Help I can’t figure out how to do zero exponents

Orlov [11]

Anything (except 0) to the 0 power is equal to 1.

8^0 = 1

(-8)^0 = 1

24781297429879741^0 = 1

4 0

2 years ago