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3 years ago

4) A mountain climber climbed up a cliff 1/4 mile at a time. he did this 5 times in one day. How many miles did he climb up? *

Mathematics

1 answer:

egoroff_w [7]3 years ago

6 0

He climbes 1 1/4 miles.
1/4+1/4+1/4+1/4+1/4= 1 1/4.

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Working together, Beth and Katy can wash a car in 6 minutes. Had she done it by herself, it would have taken Katy 18 minutes. Ho

disa [49]

Answer:

It would take Beth twelve minutes to wash the car herself

Step-by-step explanation:

If they do it together in 6 minutes.

& Katy can do it in 18 minutes alone

Then 18 - x = 6

replace x for twelve

18 - twelve = 6

Ans: twelve

7 0

3 years ago

Explain please hurry ASAP

Nuetrik [128]

Answer:

x = 4

Step-by-step explanation:

we know that 72 and (2x + 10) add up to 90

90-72 = 18

(2x + 10) = 18

2x = 8

x = 4

6 0

3 years ago

Hellp pls. This is not a test, so pls do not report.

SashulF [63]

B is the correct answer

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2 years ago

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Svet_ta [14]

Answer:

-18y = 1

y = -1/18

5x - 9(-1/18)= -2

5x + 1/2 = -4/2

5x = -5/2

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6 0

3 years ago

An industry representative claims that 10 percent of all satellite dish owners subscribe to at least one premium movie channel.

Greeley [361]

Answer: 1) 0.6561 2) 0.0037

Step-by-step explanation:

We use Binomial distribution here , where the probability of getting x success in n trials is given by :-

$P(X=x)=^nC_xp^x(1-p)^{n-x}$

, where p =Probability of getting success in each trial.

As per given , we have

The probability that any satellite dish owners subscribe to at least one premium movie channel. : p=0.10

Sample size : n= 4

Let x denotes the number of dish owners in the sample subscribes to at least one premium movie channel.

1) The probability that none of the dish owners in the sample subscribes to at least one premium movie channel = $P(X=0)=^4C_0(0.10)^0(1-0.10)^{4}$

$=(1)(0.90)^4=0.6561$

∴ The probability that none of the dish owners in the sample subscribes to at least one premium movie channel is 0.6561.

2) The probability that more than two dish owners in the sample subscribe to at least one premium movie channel.

= $P(X>2)=1-P(X\leq2)\\\\=1-[P(X=0)+P(X=1)+P(X=2)]\\\\= 1-[0.6561+^4C_1(0.10)^1(0.90)^{3}+^4C_2(0.10)^2(0.90)^{2}]\\\\=1-[0.6561+(4)(0.0729)+\dfrac{4!}{2!2!}(0.0081)]\\\\=1-[0.6561+0.2916+0.0486]\\\\=1-0.9963=0.0037$

∴ The probability that more than two dish owners in the sample subscribe to at least one premium movie channel is 0.0037.

8 0

3 years ago