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3 years ago

210 students buy milk and 196 students buy juice.How many drinks are sold that week?

Mathematics

1 answer:

lutik1710 [3]3 years ago

6 0

210 * 5 = 1050 milk

196 * 5 = 980 juice

total drinks: 2,030 drinks total

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a 60 foot tree casts a shadow 85ft long. the sine of the angle between the ground and the line that connects the tip of the shad

Jet001 [13]

The shadow to the top of the tree is 60 ft it all makes sense

8 0

3 years ago

2. Determine whether each event is impossible,

Gre4nikov [31]

Answer:

likely is the awnser

Step-by-step explanation:

likely is the awnser

3 0

3 years ago

A sample of 11 students using a Ti-89 calculator averaged 31.5 items identified, with a standard deviation of 8.35. A sample of

lina2011 [118]

Answer:

We reject H₀, with CI = 90 % we can conclude that the mean numbers of times identified with Ti-84 does not exceed that of the Ti-89 by more than 1,80

Step-by-step explanation:

Ti-89 Calculator

Sample mean x = 31,5

Sample standard deviation s₂ = 8,35

Sample size n₂ = 11

Ti-84 Calculator

Sample mean y = 46,2

Sample standard deviation s₁ = 9,99

Sample size n₁ = 12

t(s) = ( y - x - d ) / √s₁²/n₁ + s₂²/n₂

t(s) = 12,9 / √(99,8/12) + (69,72/11)

t(s) = 12,9 / √8,32 + 6,34

t(s) = 12,9 / 3,83

t(s) = 3,37

Test Hypothesis

Null Hypothesis H₀ y - x > 1,80

Altenative Hypothesis Hₐ y - x ≤ 1,80

We have a t(s) = 3,37

We need to compare with t(c) critical value for

t(c) α; n₁ +n₂-2 df = 12 +11 -2 df = 21

If we choose CI = 95 % then α = 5 % α = 0,05

From t-student table

t(c) = 1,72

t(s) = 3,37

t(s) >t(c)

t(s) is in the rejection region therefore we accept Hₐ with CI = 95 %

8 0

3 years ago

5. 81.32 times 4.6<br> Answer

Rudik [331]

374.072

Step-by-step explanation:

Use a calculator

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3 years ago

Given f: ℝ → ℝ and : ℝ → ℝ , for the following, find f(g(x)) and g(f(x)), and state the domain.

horrorfan [7]

Answer:

Remember, $(f\circ g)(x)=f(g(x))$ and the range of g must be in the domain of f.

$f(g(x))=f(x-1)=(x-1)^2-(x-1)=x^2-2x+1-x+1=x^2-3x+2$

$g(f(x))=g(x^2-x)=(x^2-x)-1=x^2-x-1$

The domain of f(g(x)) and g(f(x)) is the set of reals.

$f(g(x))=f(\sqrt{x}-2)=(\sqrt{x}-2)^2-x=\sqrt{x}^2-2*2*\sqrt{x}+2^2-x=-4\sqrt{x}+4$

$g(f(x))=g(x^2-x)=\sqrt{x^2-x}-2$

The domain of f(g(x)) is the set of nonnegative reals and the domain of g(f(x)) is the set of number such that $x^2-x\geq 0$

$f(g(x))=f(\frac{1}{x-1})=(\frac{1}{x-1})^2=\frac{1}{(x-1)^2}$

$g(f(x))=g(x^2)=\frac{1}{x^2-1}$

The domain of f(g(x)) is the set of reals except the 1 and the domain of g(f(x)) is the set of reals except the 1 and -1

$f(g(x))=f(\frac{1}{x-1})=\frac{1}{(\frac{1}{x-1}-1)}=\frac{1}{\frac{2-x}{x-1}}=\frac{x-1}{2-x}$

$g(f(x))=g(\frac{1}{x+2})=\frac{1}{(\frac{1}{x+2}-1)}=\frac{1}{\frac{-x-1}{x+2}}=\frac{x+2}{-x-1}$

The domain of f(g(x)) is the set of reals except 2, and the domain of g(f(x)) is the set of reals except -1.

$f(g(x))=f(log(2(x+3)))=f(log(2x+6))=log(2x+6)-1$

$g(f(x))=g(x-1)=log(2(x-1)+6)=log(2x+4)$

The domain of f(g(x)) is the set of nonnegative reals except -3. The domain of g(f(x)) is the set of nonnegative reals except -2.

3 0

3 years ago