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3 years ago

12

Of the 48 oranges in a carton, one-third of them were rotten. Ben and David took the remaining good oranges. If Ben took 8 more

oranges than David, how many oranges did David take?

2 answers:

vekshin13 years ago

7 0

<span>First you need to find out what is 1/3 of 48 which is 16
Then 48-16 which is 32
so now you know there are 32 oranges left
</span>David took 12 oranges and Ben took 20 oranges
12 + 20 = 32 there you go!!

kow [346]3 years ago

3 0

First you need to find out what is 1/3 of 48 which is 16
Then 48-16 which is 32
so now you know there are 32 oranges left
Ben took 8 more then David what do you think answer would be??

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Math scores on the SAT exam are normally distributed with a mean of 514 and a standard deviation of 118. If a recent test-taker

LuckyWell [14K]

Answer:

Probability that the student scored between 455 and 573 on the exam is 0.38292.

Step-by-step explanation:

We are given that Math scores on the SAT exam are normally distributed with a mean of 514 and a standard deviation of 118.

<u><em>Let X = Math scores on the SAT exam</em></u>

So, X ~ Normal( $\mu=514,\sigma^{2} =118^{2}$ )

The z score probability distribution for normal distribution is given by;

Z = $\frac{X-\mu}{\sigma}$ ~ N(0,1)

where, $\mu$ = population mean score = 514

$\sigma$ = standard deviation = 118

Now, the probability that the student scored between 455 and 573 on the exam is given by = P(455 < X < 573)

P(455 < X < 573) = P(X < 573) - P(X $\leq$ 455)

P(X < 573) = P( $\frac{X-\mu}{\sigma}$ < $\frac{573-514}{118}$ ) = P(Z < 0.50) = 0.69146

P(X $\leq$ 2.9) = P( $\frac{X-\mu}{\sigma}$ $\leq$ $\frac{455-514}{118}$ ) = P(Z $\leq$ -0.50) = 1 - P(Z < 0.50)

= 1 - 0.69146 = 0.30854

<em>The above probability is calculated by looking at the value of x = 0.50 in the z table which has an area of 0.69146.</em>

Therefore, P(455 < X < 573) = 0.69146 - 0.30854 = <u>0.38292</u>

Hence, probability that the student scored between 455 and 573 on the exam is 0.38292.

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3 years ago

2 3/4 divided by 1/2 ?

Vladimir [108]

Answer:

5.5

Step-by-step explanation:

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2 years ago

(5 points) An urn contains two blue balls denoted by B1 and B2, and three white balls denoted by W1, W2 and W3. One ball is draw

never [62]

Answer:

The probability of the event that first ball that is drawn is blue is $\frac 25$ .

Step-by-step explanation:

Probability:

If S is is an sample space in which all outcomes are equally likely and E is an event in S, then the probability of E,denoted P(E) is

$P(E)=\frac{\textrm{The number of outcomes E}}{\textrm{The total number outcomes of S}}$

Given that,

An urn contains two balls B₁ and B₂ which are blue in color and W₁,W₂ and W₃ which are white in color.

Total number of ball =(2+3) =5

The number ways of selection 2 ball out of 5 ball is

=5²

=25

Total outcomes = 25

List of all outcomes in the event that the first ball that is drawn is blue are

B₁B₁ , B₁B₂ , B₁W₁ , B₁W₂ , B₁W₃ , B₂B₁ , B₂B₂ , B₂W₁ , B₂W₂ , B₂W₃

The number of event that the first ball that is drawn is blue is

=10.

The probability of the event that first ball that is drawn is blue is

$=\frac{10}{25}$

$=\frac25$

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3 years ago

Help, please giving 20 points

scZoUnD [109]

Answer:

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Step-by-step explanation:

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2 years ago

Recent census data indicated that 14.2% of adults between the ages of 25 and 34 live with their parents. A random sample of 125

kicyunya [14]

Answer:

The probability is $P(14 < X < 20 ) = 0.5354$

Step-by-step explanation:

From the question we are told that

The proportion that live with their parents is $\r p = 0.142$

The sample size is n = 125

Given that there are two possible outcomes and that this outcomes are independent of each other then we can say the Recent census data follows a Binomial distribution

i.e

$X \ \~ \ B( \mu , \sigma )$

Now the mean is evaluated as

$\mu = n * \r p$

$\mu = 125 * 0.142$

$\mu = 17.75$

Generally the proportion that are not staying with parents is

$\r q = 1 - \r p$

= > $\r q = 0.858$

The standard deviation is mathematically evaluated as

$\sigma = \sqrt{n * \r p * \r q }$

$\sigma = \sqrt{ 125 * 0.142 * 0.858 }$

$\sigma = 3.90$

Given the n is large then we can use normal approximation to evaluate the probability as follows

$P(14 < X < 20 ) = P( \frac{ 14 - 17.75}{3.90}$

Now applying continuity correction

$P(14 < X < 20 ) = P( \frac{ 13.5 - 17.75}{3.90} < \frac{ X - \mu }{\sigma } < \frac{ 19.5 - 17.75}{3.90} )$

Generally

$\frac{ X - \mu }{\sigma } = Z ( The \ standardized \ value \ of X )$

$P(14 < X < 20 ) = P( \frac{ 13.5 - 17.75}{3.90} < Z< \frac{ 19.5 - 17.75}{3.90} )$

$P(14 < X < 20 ) = P( -1.0897 < Z< 0.449 } )$

$P(14 < X < 20 ) = P( Z< 0.449 ) - P(Z < -1.0897)$

So for the z - table

$P( Z< 0.449 ) = 0.67328$

$P(Z < -1.0897) = 0.13792$

$P(14 < X < 20 ) = 0.67328 - 0.13792$

$P(14 < X < 20 ) = 0.5354$

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3 years ago