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3 years ago

12

A gas tank is 1/4 full. When Raoul puts 4 gallons of gasoline into the tank, it is 5/8 full. How many gallons does a full tank h

old?

2 answers:

zaharov [31]3 years ago

6 0

Answer:

i think it holds 10.6 gallons

Step-by-step explanation:

i changed 1/4 to have a common denominator with 5/8

so now 1/4 is 2/8

i figured out that 4 gallons of gas fills the tank up 3/8 way

1/8 of the tank full equals 1.333 (continuous)

6/8 of the tank full is 8 gallons plus 2/8 is 2.6 more gallons add 8 and 2.6

you get 10.6 gallons to fill the tank.

disa [49]3 years ago

4 0

Isn’t it 8 gallons because it’s 5/8 full?

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Find the quotient: –10/19÷(−5/7)

Semmy [17]

Answer:

14/19

Step-by-step explanation:

Note that there are two " - " signs here. Thus, the end result will be " + "

We move from left to right, obtaining first 10/19 divided by (5/7):

10

----- ÷ (5/7)

19

To divide by (5/7), invert (5/7) and then multiply. We get:

10 7 2(7)

----- * ----- = ------- = 14/19

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3 years ago

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3 years ago

Read 2 more answers

In a process that manufactures bearings, 90% of the bearings meet a thickness specification. A shipment contains 500 bearings. A

Marina86 [1]

Answer:

(a) 0.94

(b) 0.20

(c) 90.53%

Step-by-step explanation:

From a population (Bernoulli population), 90% of the bearings meet a thickness specification, let $p_1$ be the probability that a bearing meets the specification.

So, $p_1=0.9$

Sample size, $n_1=500$ , is large.

Let X represent the number of acceptable bearing.

Convert this to a normal distribution,

Mean: $\mu_1=n_1p_1=500\times0.9=450$

Variance: $\sigma_1^2=n_1p_1(1-p_1)=500\times0.9\times0.1=45$

$\Rightarrow \sigma_1 =\sqrt{45}=6.71$

(a) A shipment is acceptable if at least 440 of the 500 bearings meet the specification.

So, $X\geq 440.$

Here, 440 is included, so, by using the continuity correction, take x=439.5 to compute z score for the normal distribution.

$z=\frac{x-\mu}{\sigma}=\frac{339.5-450}{6.71}=-1.56$ .

So, the probability that a given shipment is acceptable is

$P(z\geq-1.56)=\int_{-1.56}^{\infty}\frac{1}{\sqrt{2\pi}}e^{\frac{-z^2}{2}}=0.94062$

Hence, the probability that a given shipment is acceptable is 0.94.

(b) We have the probability of acceptability of one shipment 0.94, which is same for each shipment, so here the number of shipments is a Binomial population.

Denote the probability od acceptance of a shipment by $p_2$ .

$p_2=0.94$

The total number of shipment, i.e sample size, $n_2= 300$

Here, the sample size is sufficiently large to approximate it as a normal distribution, for which mean, $\mu_2$ , and variance, $\sigma_2^2$ .

Mean: $\mu_2=n_2p_2=300\times0.94=282$

Variance: $\sigma_2^2=n_2p_2(1-p_2)=300\times0.94(1-0.94)=16.92$

$\Rightarrow \sigma_2=\sqrt(16.92}=4.11$ .

In this case, X>285, so, by using the continuity correction, take x=285.5 to compute z score for the normal distribution.

$z=\frac{x-\mu}{\sigma}=\frac{285.5-282}{4.11}=0.85$ .

So, the probability that a given shipment is acceptable is

$P(z\geq0.85)=\int_{0.85}^{\infty}\frac{1}{\sqrt{2\pi}}e^{\frac{-z^2}{2}=0.1977$

Hence, the probability that a given shipment is acceptable is 0.20.

(c) For the acceptance of 99% shipment of in the total shipment of 300 (sample size).

The area right to the z-score=0.99

and the area left to the z-score is 1-0.99=0.001.

For this value, the value of z-score is -3.09 (from the z-score table)

Let, $\alpha$ be the required probability of acceptance of one shipment.

So,

$-3.09=\frac{285.5-300\alpha}{\sqrt{300 \alpha(1-\alpha)}}$

On solving

$\alpha= 0.977896$

Again, the probability of acceptance of one shipment, $\alpha$ , depends on the probability of meeting the thickness specification of one bearing.

For this case,

The area right to the z-score=0.97790

and the area left to the z-score is 1-0.97790=0.0221.

The value of z-score is -2.01 (from the z-score table)

Let $p$ be the probability that one bearing meets the specification. So

$-2.01=\frac{439.5-500 p}{\sqrt{500 p(1-p)}}$

On solving

$p=0.9053$

Hence, 90.53% of the bearings meet a thickness specification so that 99% of the shipments are acceptable.

8 0

4 years ago

Select the correct answer.

Elan Coil [88]

We can make a conclusion about the data collection method used was fair since both shoe designs were tested on identical parameters option (B) is correct.

<h3>What is simple random sampling?</h3>

Simple random sampling is a sampling strategy in which every item in the population has an equal probability of being chosen for the sample.

We have:

Two manufacturers of the shoe company:

Fastex Shoes
SportsPlus

SportsPlus came up with a design that it claims performs even better than Fastex's design.

SportsPlus decided to test the claims:

It created two groups of 15 randomly selected athletes each.
Members of both groups were pretested and determined to have equal athletic ability.
Additionally, they followed the same nutrition plan and training program during the study.

Based on the data the statement "The data collection method used was fair since both shoe designs were tested on identical parameters".

Thus, we can make a conclusion about the data collection method used was fair since both shoe designs were tested on identical parameters option (B) is correct.

Learn more about the simple random sampling here:

brainly.com/question/13219833

#SPJ1

3 0

2 years ago

WILL MARK BRAINLIEST

alexandr1967 [171]

Answer:

Equation: 7x - 23 = 25 + x

Solution: x = 8

Step-by-step explanation:

7x - 23 = 25 + x

1. Subtract 25 from both sides:

7x - 23 - 25 = 25 - 25 + x

7x - 48 = x

2. Subtract 7x from both sides to get x on one side:

7x - 7x - 48 = x - 7x

- 48 = -6x

3. Divide both sides by -6 to get x by itself:

8 = x

8 is the solution

4 0

3 years ago