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2 years ago

Emma bought three cans of tennis balls for $4 each. Sales tax for all three cans was $0.95. Wh

Mathematics

2 answers:

LiRa [457]2 years ago

6 0

Answer:

4(3)+0.95=X

Step-by-step explanation:

pav-90 [236]2 years ago

3 0

Answer:

A sixth-grade student bought three cans of tennis balls for $4 each.

The sales tax for each can was $0.95.

To Find :

The total amount the student spent.

Solution :

Price of each can including taxes :

p = $( 4 + 0.95 )

p = $4.95

So, price of 3 such cans including tax is :

P = p×3

P = $( 4.95×3 )

P = $14.85

Therefore, total amount spend by the student is $14.85 .

Step-by-step explanation:

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Question:

What is an equation of the line that passes through the point (3,6) and is parallel to the line 4x-3y=21

Answer:

The equation of the line that passes through the point (3,6) and is parallel to the line 4x-3y=21 is:

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Solution:

Given that line passes through the point (3,6) and is parallel to the line 4x-3y=21

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$m = \frac{4}{3}$

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Substitute m = 4/3 and (x, y) = (3, 6) in eqn 1

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Step-by-step explanation:

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This question is incomplete, the complete question;

A highway department executive claims that the number of fatal accidents which occur in her state does not vary from month to month. The results of a study of 193 fatal accidents were recorded. Is there enough evidence to reject the highway department executive's claim about the distribution of fatal accidents between each month? Month Jan Feb Mar Apr May Jun Jul Aug Sep Oct Nov Dec Fatal Accidents 11 13 22 11 15 29 15 9 13 15 22 18

Step 10 of 10 : State the conclusion of the hypothesis test at the 0.1 level of significance.

Answer: H₀ : The distribution of fatal accidents between each month is same

Step-by-step explanation:

H₀ : The distribution of fatal accidents between each month is same

H₁ : The distribution of fatal accidents between each month is different

Let the los be alpha 0.10

given data ;

Observed Expected

Mon Freq (Oi) Freq Ei (Oi-Ei)^2 /Ei

Jan 11 16.0833 1.606649

Feb 13 16.0833 0.591105

Mar 22 16.0833 2.176598

Apr 11 16.0833 1.606649

May 15 16.0833 0.072971

Jun 29 16.0833 10.373489

Jul 15 16.0833 0.072971

Aug 9 16.0833 3.119603

Sep 13 16.0833 0.591105

Oct 15 16.0833 0.072971

Nov 22 16.0833 2.176598

Dec 18 16.0833 0.228411

Total: 193 93 22.689119

Expected Freq ⇒ 193 / 12 = 16.08333

Test Statistic, x² : 22.6891

Num Categories: 12

Degrees of freedom: 12 - 1 = 11

Critical value X² : 24.725

P-Value: 0.0195

since Chi-square value < Chi-square critical value

and P-value > alpha 0.01 so we accept H₀

Therefore we conclude that the distribution of fatal accidents between each month is same

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