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4 years ago

PLEASE HELP ASAP!!

Mathematics

1 answer:

Art [367]4 years ago

7 0

Answer:

See explanation

Step-by-step explanation:

A. P(top)=top outcomes/all kinds of outcomes=4/30=2/15=13.33333333333...%

P(bottom)=bottom outcomes/ all kinds of outcomes=1/30=3.33333333....%

P(side)=25/30=5/6=83.33333333333...%

B. No. If were equally likely, the probabilities for A would have been roughly the same. It seems like the side event is more probable.

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PLZ HELP!!!! WILL MARK AS BRANLESTTT.. PLZZZZ

Iteru [2.4K]

Answer:

(4,4)

Step-by-step explanation:

1+3=4

3+1=4

(4,4)

7 0

3 years ago

there arre 160 boys and girls playing in the soccer game 32 of students are wearing orange what %of the players are wearing oran

mash [69]

Answer:

20%

Step-by-step explanation:

To find this, set up a ratio of percent over 100 equals is over of. Is being 32 and of being 160.

$\frac{32}{160}$ = $\frac{x}{100}$

Then, cross multiply. Multiply 32 by 100 before dividing by 160 in order to get the final answer of 20%.

32 x 100 = 3200

3200/160 = 20

Hope this helps!

8 0

4 years ago

(True or False) The Long Division symbol is the same as the Radical (Root) symbol?

Andrews [41]

Answer:

True.

Step-by-step explanation:

4 0

3 years ago

A tire manufacturer wants to estimate the average number of miles that may be driven in a tire of a certain type before the tire

alexandr402 [8]

Answer:

97% confidence interval for the average number of miles that may be driven is [26.78 miles, 33.72 miles].

Step-by-step explanation:

We are given that a random sample of tires is chosen and are driven until they wear out and the number of thousands of miles is recorded;

32, 33, 28, 37, 29, 30, 22, 35, 23, 28, 30, 36.

Firstly, the pivotal quantity for finding the confidence interval for the population mean is given by;

P.Q. = $\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }$ ~ $t_n_-_1$

where, $\bar X$ = sample average number of miles = $\frac{\sum X}{n}$ = 30.25

s = sample standard deviation = $\sqrt{\frac{\sum (X-\bar X)^{2} }{n-1} }$ = 4.71

n = sample of tires = 12

$\mu$ = population average number of miles

Here for constructing a 97% confidence interval we have used One-sample t-test statistics as we don't know about population standard deviation.

So, 97% confidence interval for the population mean, $\mu$ is ;

P(-2.55 < $t_1_1$ < 2.55) = 0.97 {As the critical value of t at 11 degrees of

freedom are -2.55 & 2.55 with P = 1.5%}

P(-2.55 < $\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }$ < 2.55) = 0.97

P( $-2.55 \times {\frac{s}{\sqrt{n} } }$ < ${\bar X-\mu}$ < $2.55 \times {\frac{s}{\sqrt{n} } }$ ) = 0.97

P( $\bar X-2.55 \times {\frac{s}{\sqrt{n} } }$ < $\mu$ < $\bar X+2.55 \times {\frac{s}{\sqrt{n} } }$ ) = 0.97

97% confidence interval for $\mu$ = [ $\bar X-2.55 \times {\frac{s}{\sqrt{n} } }$ , $\bar X+2.55 \times {\frac{s}{\sqrt{n} } }$ ]

= [ $30.25-2.55 \times {\frac{4.71}{\sqrt{12} } }$ , $30.25+2.55 \times {\frac{4.71}{\sqrt{12} } }$ ]

= [26.78 miles, 33.72 miles]

Therefore, 97% confidence interval for the average number of miles that may be driven is [26.78 miles, 33.72 miles].

7 0

4 years ago

What is the answer for this problem

Margarita [4]

The answer is 2 and negative 9

3 0

3 years ago