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3 years ago

2. Draw a right triangle with side lengths of 3, 4, and 5 units.

Mathematics

1 answer:

puteri [66]3 years ago

7 0

See the attached picture:

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41,000.00 minus 23,000.00

inna [77]

18,000.00.This is the answer

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3 years ago

A map that has a scale of 3 cm: 18 km. If Riverside and Smithville are 54km apart then they are how far apart on the map?

SVETLANKA909090 [29]

Setting up the ratio, you get 3/18=x/54. Solve the proportion, you get that Riverside and Smithville are 9 cm apart on the map.

8 0

2 years ago

Please answer correctly I will mark you Brainliest!

BartSMP [9]

Answer:

1047.12in³

Step-by-step explanation:

The volume of the sphere is:

V= 4/3 πr³

Where r is radius

Therefore the volume of each pinata is:

V= 4/3 x π x 5³

V= 523.6in³

The total:

V = 523.6 + 523.6 = 1047.12in³

7 0

3 years ago

Find the product. (x - 4)²

gulaghasi [49]

Answer:

x^2−8x+16

Step-by-step explanation:

5 0

3 years ago

Read 2 more answers

A person drilled a hole in a die and filled it with a lead weight, then proceeded to roll it 200 times. Here are the observed f

Alona [7]

Solution :

The objective of the study is to test the claim that the loaded die behaves at a different way than a fair die.

Null hypothesis, $H_0:p_1=p_1=p_3=p_4=p_5=p_6=\frac{1}{6}$

That is the loaded die behaves as a fair die.

Alternative hypothesis, $H_a$ : loaded die behave differently than the fair die.

Number of attempts , n = 200

Expected frequency, $E_i=np_i$

$=200 \times \frac{1}{6} = 33.333$

Test statistics, $x^2= \sum^6_{i=1} \frac{(O_i-E_i)^2}{E_i}$

$=\frac{(28-33.333)^2}{33.333}+\frac{(29-33.333)^2}{33.333}+\frac{(40-33.333)^2}{33.333}+\frac{(41-33.333)^2}{33.333}+\frac{(28-33.333)^2}{33.333}+$ $\frac{(34-33.333)^2}{33.333}$

≈ 5.8

Degrees of freedom, df = n - 1

= 6 - 1

= 5

Level of significance, α = 0.10

At α = 0.10 with df = 5, the critical value from the chi square table

$x^2_{\alpha}= \text{chi inv}(0.10,5)$

= 9.236

Thus the critical value is $x_{\alpha}^2=9.236$

$P \text{ value} = P[x^2_{df} \geq x^2]$

$=P[x^2_5\geq 5.80]$

= chi dist (5.80, 5)

= 0.3262

Decision : The value of test statistics 5.80 is not greater than the critical value 9.236, thus fail to reject $H_o$ at 10% LOS.

Conclusion : There is no enough evidence to support the claim that the loaded die behave in a different than a fair die.

3 0

3 years ago