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Estimate the total cost of 31 televisions at £196.50 each and 19 DVD players at £50.99 each

Gwar [14]

Answer:

Solution,

Cost of one television=£196.50

cost of 31 televisions:

 $= 196.50 \times 31 \\ = 6091.5$ 

Cost of one DVD player=£50.99

Cost of 19 DVD player:

 $= 50.99 \times 19 \\ = 968.81$ 

Now,

Total cost

= cost of 31 televisions + cost of 19 DVD player

=6091.5+968.81

=£7060.31

Hope this helps...

Good luck on your assignment..

3 0

4 years ago

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Which is the graph of y = 5log(x+3)

Maurinko [17]

Answer:

We are given a equation as:

5log(x+3)=5

We are asked to find a graph that is used to solve the above equation.

We can write the given equation as:

we will divide both side of the equation by 5 to obtain:

log(x+3)=1

Now we have to determine which graph represents the function:

y=log(x+3)

since we know that when x=-2.

y=log(-2+3)=log(1)=0

Hence, the graph should pass through (-2,0).

Hence, the graph that satisfies this is attached to the answer.

Step-by-step explanation:

7 0

3 years ago

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If it's 2:30 PM what time is it gonna be in 45 minutes

natta225 [31]

3:15pm hope this helps bff

7 0

3 years ago

To test the effect of classical music on the brain, a study has been done. Twenty 6th grade students are randomly divided into t

Volgvan

Answer:

Step-by-step explanation:

Hello!

A study was conducted to test the effect of classical music on the brain. For this 20 6th grade students were randomly divided into two independent groups of 10, the same math test was given to these students. The first group listened to classical music for 20 min before taking the test. The second group took the test without listening to music.

Sample 1 (With music)

X₁: Score of a 6th grade student that heard 20 min classical music before taking the math test.

91 77 58 89 83 78 74 81 91 88

n₁= 10

X[bar]₁= 81

S₁= 10.11

Sample 2 (Without music)

X₂: Score of a 6th grade student that didn't hear classical music before taking the math test.

81 65 69 69 67 61 67 87 64 81

n₂= 10

X[bar]₂= 71.10

S₂= 8.70

Asuming both variables have a normal distribution and the population variances are unknown but equal, the statistic to use for both the CI and hypothesis tests is:

t= (X[bar]₁-X[bar]₂) - (μ₁ - μ₂) ~t $_{n_1+n_2-2}$

Sa $\sqrt{\frac{1}{n_1} + \frac{1}{n_2} }$

a) 95% for (μ₁ - μ₂)

(X[bar]₁-X[bar]₂) ± $t_{n_1+n_2-2; 1-\alpha /2}$ *Sa $\sqrt{\frac{1}{n_1} + \frac{1}{n_2} }$

$Sa^2= \frac{(n_1-1)S_1^2+(n_2-1)S_2^2)}{n_1+n_2-2}$

$Sa^2= \frac{9*102.22+9*75.66}{18}$

Sa²= 88.94

Sa= 9.4308 ≅ 9.43

$t_{n_1+n_2-2;1-\alpha /2} = t_{18; 0.975} = 2.101$

(81-71.10) ± 2.101*(9.43* $\sqrt{\frac{1}{10} + \frac{1}{10} }$ )

[1.04;18.76]

With a confidence level of 95% youd expect that the interval [1.04;18.76] will contain the difference between the population means of the test scores of the kids that listened to classic music and the kids that didn't listen to music before taking the math test.

b)

H₀: μ₁ ≤ μ₂

H₁: μ₁ > μ₂

α: 0.05

One-tailed test (right tail)

Critical value

$t_{n_1+n_2-2; 1 - \alpha } = t_{18; 0.95} = 1.734$

Rejection region t ≥ 1.734

t= (81-71.10) - 0 = 2.34

9.43* $\sqrt{\frac{1}{10} + \frac{1}{10} }$

The decision is to reject the null hypothesis.

c) You are asked to conduct the same test at a different levelm this means that only the significance level changes:

H₀: μ₁ ≤ μ₂

H₁: μ₁ > μ₂

α: 0.01

One-tailed test (right tail)

Critical value

$t_{n_1+n_2-2; 1 - \alpha } = t_{18; 0.99} = 2.552$

Rejection region t ≥ 2.552

t= (81-71.10) - 0 = 2.34

9.43* $\sqrt{\frac{1}{10} + \frac{1}{10} }$

The decision is to not reject the null hypothesis.

At a significance level of 5%, the decision is to reject the null hypothesis, which means that the population average of the test scores of 6th-grade children that listened to classical music before taking a math test is greater than the population average of the test scores of 6th graders that took the math test without listening to classical music.

But at 1% significance level, there is not enough evidence to reject the null hypothesis. At this level, the conclusion is that the average test score of 6th graders that listened to classical music before taking the math test is at most equal to the average test score of 6th graders that didn't listen to music before the test.

I hope it helps!

4 0

3 years ago

There are 48 calories of energy in 40g of rice. How many calories are in 160 g of rice?

Rashid [163]

160 grams of rice is four times larger than the given 40 g.
So the number of Calories must be 4 times larger too. 

48 x 4 = 192 Calories in 160 g of rice

5 0

3 years ago

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