Find the sum of -38 + (-57)

Jannet saved 22 dollars one month and 39 dollars the next month. She wants to by a bicycle that costs $ 100. About how much more

coldgirl [10]

22 dollars + 39 dollars = 61 dollars

100-61= 39 dollars. She needs $39 more

7 0

3 years ago

A salad bar offers vegetable platters in 4 sizes (small, medium, large, and super-sized) and 6 different vegetables to choose fr

quester [9]

Answer: She could create 60 platters.

Step-by-step explanation:

Given: A salad bar offers vegetable platters in 4 sizes (small, medium, large, and super-sized) and 6 different vegetables to choose from.

Number of combinations of selecting <em>r</em> things from <em>n</em> things is given by:-

$^nC_r=\dfrac{n!}{r!(n-r)!}$

So, the number of combinations of selecting 2 different vegetables from 6 = $^6C_2=\dfrac{6!}{2!4!}=\dfrac{6\times5}{2}=15$

Now, By Fundamental counting principle, the number of different platters she could create = (number of ways of selecting 2 different vegetables from 6) x (Number of sizes)

= 15 x 4

= 60

Hence, she could create 60 platters.

6 0

3 years ago

Can someone help me find the equivalent expressions to the picture below? I’m having trouble

miss Akunina [59]

Answer:

Options (1), (2), (3) and (7)

Step-by-step explanation:

Given expression is $\frac{\sqrt[3]{8^{\frac{1}{3}}\times 3} }{3\times2^{\frac{1}{9}}}$ .

Now we will solve this expression with the help of law of exponents.

$\frac{\sqrt[3]{8^{\frac{1}{3}}\times 3} }{3\times2^{\frac{1}{9}}}=\frac{\sqrt[3]{(2^3)^{\frac{1}{3}}\times 3} }{3\times2^{\frac{1}{9}}}$

$=\frac{\sqrt[3]{2\times 3} }{3\times2^{\frac{1}{9}}}$

$=\frac{2^{\frac{1}{3}}\times 3^{\frac{1}{3}}}{3\times 2^{\frac{1}{9}}}$

$=2^{\frac{1}{3}}\times 3^{\frac{1}{3}}\times 2^{-\frac{1}{9}}\times 3^{-1}$

$=2^{\frac{1}{3}-\frac{1}{9}}\times 3^{\frac{1}{3}-1}$

$=2^{\frac{3-1}{9}}\times 3^{\frac{1-3}{3}}$

$=2^{\frac{2}{9}}\times 3^{-\frac{2}{3} }$ [Option 2]

$2^{\frac{2}{9}}\times 3^{-\frac{2}{3} }=(\sqrt[9]{2})^2\times (\sqrt[3]{\frac{1}{3} } )^2$ [Option 1]

$2^{\frac{2}{9}}\times 3^{-\frac{2}{3} }=(\sqrt[9]{2})^2\times (\sqrt[3]{\frac{1}{3} } )^2$

$=(2^2)^{\frac{1}{9}}\times (3^2)^{-\frac{1}{3} }$

$=\sqrt[9]{4}\times \sqrt[3]{\frac{1}{9} }$ [Option 3]

$2^{\frac{2}{9}}\times 3^{-\frac{2}{3} }=(2^2)^{\frac{1}{9}}\times (3^{-2})^{\frac{1}{3} }$

$=\sqrt[9]{2^2}\times \sqrt[3]{3^{-2}}$ [Option 7]

Therefore, Options (1), (2), (3) and (7) are the correct options.

6 0

2 years ago

a 5 sided solid has the numbers 1,2,3,4, and 5. what is the probability of rolling two five-sided solids and getting a sum of ei

kipiarov [429]

So this is going to be alot of writing to show my thinking but ill bold the answer.

1,1
1,2
1,3
1,4
1,5

2,1
2,2
2,3
2,4
2,5

3,1
3,2
3,3
3,4
3,5

4,1
4,2
4,3
4,4
4,5

5,1
5,2
5,3
5,4
5,5

next ill mark all the ones that equal 4 or 8 when added together, with an x

1,1
1,2
x1,3
1,4
1,5

2,1
x2,2
2,3
2,4
2,5

x3,1
3,2
3,3
3,4
x3,5

4,1
4,2
4,3
x4,4
4,5

5,1
5,2
x5,3
5,4
5,5

that is 6 (that equal 4 or 8) out of 25

so your ratio would be 6:19

6 0

3 years ago

The world population at the beginning of 1980 was 4.5 billion. Assuming that the population continued to grow at the rate of app

AfilCa [17]

Answer:

$Q(t) = 4.5(1.013)^{t}$

The world population at the beginning of 2019 will be of 7.45 billion people.

Step-by-step explanation:

The world population can be modeled by the following equation.

$Q(t) = Q(0)(1+r)^{t}$

In which Q(t) is the population in t years after 1980, in billions, Q(0) is the initial population and r is the growth rate.

The world population at the beginning of 1980 was 4.5 billion. Assuming that the population continued to grow at the rate of approximately 1.3%/year.

This means that $Q(0) = 4.5, r = 0.013$

So

$Q(t) = Q(0)(1+r)^{t}$

$Q(t) = 4.5(1.013)^{t}$

What will the world population be at the beginning of 2019 ?

2019 - 1980 = 39. So this is Q(39).

$Q(t) = 4.5(1.013)^{t}$

$Q(39) = 4.5(1.013)^{39} = 7.45$

The world population at the beginning of 2019 will be of 7.45 billion people.

6 0

3 years ago