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3 years ago

How much would each starburst weigh if 3 starburst weigh 1 ounce?

Mathematics

1 answer:

sineoko [7]3 years ago

6 0

Answer:

3 ounces

Step-by-step explanation:

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For this problem, involving a weighted die, assume that the outcomes 1 through 4 are equally likely, that the outcomes 5 through

storchak [24]

Answer:

a. $\frac{3}{14}$

b. $\frac{1}{14}$

Step-by-step explanation:

We are given that a die is rolled

Let x bet the cases in favor of 5.

Number of cases in favor of 6=x

Then according to question

Number of cases in favor of 1=3x

Number of cases in favor of 2=3x

Number of cases in favor of 3=3x

Number of cases in favor of 4=3x

Sum of total cases= $x+x+3x+3x+3x+3x=14x$

Probability of getting 5 = $\frac{x}{14x}=\frac{1}{14}$

Probability of getting 1= $\frac{3x}{14x}=\frac{3}{14}$

1.We have to find probability should be assigned to rolling a 4.

The probability of getting = $\frac{3}{14}$

2.We have to find the probability of rolling 6.

The probability of getting 6= $\frac{1}{14}$

3 0

4 years ago

The logarithm of the base to it self is always......

Elanso [62]

Answer:

Step-by-step explanation:

Logₐa = 1

6 0

3 years ago

Read 2 more answers

An angle measures 64.8° less than the measure of its complementary angle. What is the measure of each angle?

kirill115 [55]

Answer:

77.4 degrees and 12.6 degrees

Step-by-step explanation:

Let the angles be x and x - 64.8.

Therefore, on the basis of knowledge that complementary angles add up to 90 degrees :

x + x - 64.8 = 90

2x = 154.8

x = 77.4 degrees

x - 64.8 = 77.4 - 64.8 = 12.6 degrees

5 0

3 years ago

Which of the following is incorrect?

jeyben [28]

C is wrong...............

8 0

3 years ago

Read 2 more answers

If 1200 cm2 of material is available to make a box with a square base and an open top, find the largest possible volume of the b

Colt1911 [192]

Answer:

The largest possible volume V is ;

V = l^2 × h

V = 20^2 × 10 = 4000cm^3

Step-by-step explanation:

Given

Volume of a box = length × breadth × height= l×b×h

In this case the box have a square base. i.e l=b

Volume V = l^2 × h

The surface area of a square box

S = 2(lb+lh+bh)

S = 2(l^2 + lh + lh) since l=b

S = 2(l^2 + 2lh)

Given that the box is open top.

S = l^2 + 4lh

And Surface Area of the box is 1200cm^2

1200 = l^2 + 4lh ....1

Making h the subject of formula

h = (1200 - l^2)/4l .....2

Volume is given as

V = l^2 × h

V = l^2 ×(1200 - l^2)/4l

V = (1200l - l^3)/4

the maximum point is at dV/dl = 0

dV/dl = (1200 - 3l^2)/4

dV/dl = (1200 - 3l^2)/4 = 0

3l^2= 1200

l^2 = 1200/3 = 400

l = √400

I = 20cm

Since,

h = (1200 - l^2)/4l

h = (1200 - 20^2)/4×20

h = (800)/80

h = 10cm

The largest possible volume V is ;

V = l^2 × h

V = 20^2 × 10 = 4000cm^3

4 0

3 years ago