By rounding to 1 significant figure, estimate the value of 48.7 x 61.2<br> 11.3

Vera_Pavlovna [14]

Answer:

The answer to this problem is True.

7 0

3 years ago

The regular price of a burger meal at a certain restaurant is $8.70. It is on sale for a 20% discount. What is the sale price of

seropon [69]

Answer:

1.74

Step-by-step explanation:

4 0

2 years ago

Suppose a shipment of 120 electronic components contains 3 defective components. to determine whether the shipment should be ac

Likurg_2 [28]

For this problem, the most accurate is to use combinations

Because the order in which it was selected in the components does not matter to us, we use combinations

Then the combinations are $nC_r = \frac{ n! }{r! (n-r)!}$

n represents the amount of things you can choose and choose r from them

You need the probability that the 3 selected components at least one are defective.

That is the same as:

(1 - probability that no component of the selection is defective).

The probability that none of the 3 selected components are defective is:

$P = \frac{_{117}C_3}{_{120}C_3}$

Where $_{117}C_3$ is the number of ways to select 3 non-defective components from 117 non-defective components and $_{120}C_3$ is the number of ways to select 3 components from 120.

$_{117}C_3 = 260130$

$_{120}C_3 = 280840$

So:

$P = \frac{260130}{280840} = 0.927$

Finally, the probability that at least one of the selected components is defective is:

$P = 1-0.927 = 0.0737$

P = 7.4%

3 0

3 years ago

Two surfers and statistics students collected data on the number of days on which surfers surfed in the last month for 30 longbo

Alisiya [41]

Answer:

Do not reject H0. The mean days surfed for longboarders is significantly larger than the mean days surfed for all shortboarders

Step-by-step explanation:

The null hypothesis is that the mean days surfed for all long boarders is larger than the mean days surfed for all short boarders

H0: μL > μs against the claim Ha: μL≤ μs

the alternate hypothesis is the mean days surfed for all long boarders isless or equal to the mean days surfed for all short boarders (because long boards can go out in many different surfing conditions)

The test statistic is

t= x1- x2/ √s1/n1+ s2/n2

1) Calculations

Longboards

Mean

ˉx=∑x/n=4+8+9+4+9+7+9+6+6+11+15+13+16+12+10+12+18+20+15+10+15+19+21+9+22+19+23+13+12+10/30

=377/30

=12.5667

Longboard Variance S2=[∑dx²-(∑dx)²/n]/n-1

=[831-(-13)²/30]/29

=831-5.6333/29

=825.3667/29

=28.4609

Shortboard Mean

ˉx=∑x/n=6+4+6+6+7+7+7+10+4+6+7+5+8+9+4+15+13+9+12+11+12+13+9+11+13+15+9+19+20+11/30

=288/30

=9.6

Shortboard Variance S2=[∑x²-(∑x)²/n]/n-1

=[ 3270-(288)2/30]/29

=3270-2764.8/29

=505.2/29

=17.4207

2) Putting values in the test statistic

t=|x1-x2|/√S²1/n1+S²2/n2

t =|12.5667-9.6|/√28.4609/30+17.4207/30

t =|2.9667|/√0.9487+0.5807

t=|2.9667|/√1.5294

t=|2.9667|/1.2367

t=2.3989

3) Degree of freedom =n1+n2-2=30+30-2=58

4) The critical region is t ≤ t(0.05) (58) =1.6716

5) Since the calculated t= 2.4 does not fall in the critical region t(0.05) (58) ≤ 1.6716 we do not reject H0.

The p-value is 0.008969. The result is significant at p <0 .05.

6 0

3 years ago

A worker is handling the four of a rectangle room that is 12 feet by 15 feet.The tiles are square with side lengths 1 1/3 feet.

Anon25 [30]

Answer:

Step-by-step explanation:

The measure of the floor of the rectangular room that is 12 feet by 15 feet. The formula for determining the area of a rectangle is expressed as

Area = length × width

Area of the rectangular room would be

12 × 15 = 180 feet²

The tiles are square with side lengths 1 1/3 feet. Converting 1 1/3 feet to improper fraction, it becomes 4/3 feet

Area if each tile is

4/3 × 4/3 = 16/9 ft²

The number of tiles needed to cover the entire floor is

180/(16/9) = 180 × 9/16

= 101.25

102 tiles would be needed because the tiles must be whole numbers.

3 0

2 years ago

By rounding to 1 significant figure, estimate the value of 48.7 x 61.2 11.3