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Advocard [28]
3 years ago
11

Natasha and Richard win some money and share it in the ratio 5:2. Natasha gets £75. How much did Richard get?

Mathematics
1 answer:
Evgesh-ka [11]3 years ago
7 0

Answer:

Richard got $30

Step-by-step explanation:

To find how much money Richard got, use cross multiplication:

\frac{5}{2}=\frac{75}{x}

5x = 150

x = 30

Richard got $30.

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1.875

Step-by-step explanation:

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Evaluate the expression below when y = -3.
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2y+7
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Using the following diagram, solve for X.
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3 years ago
I need Help plsssss!!!!!
laiz [17]

Answer:

2x - 3

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Step-by-step explanation:

The root of a problem is the x value that makes the equation equal to zero.

If the factored form of the equation is (2x-3)(x-9) that means if either 2x - 3 or x - 9 is equal to zero, the entire equation will equal zero, because zero times any number is equal to zero.  

Therefore you must solve for (2x-3)=0 and (x-9)=0

4 0
3 years ago
According to the U.S. Bureau of the Census, the average age of brides marrying for the first time is 23.9 years with a populatio
JulsSmile [24]

Answer:

We conclude that young women are delaying marriage and marrying at a later age.

Step-by-step explanation:

We are given that the average age of brides marrying for the first time is 23.9 years with a population standard deviation of 4.2 years.

The sociologist randomly samples 100 marriage records and determines the average age of the first time brides is 24.9 years.

Let \mu = <u><em>average age of brides marrying for the first time.</em></u>

So, Null Hypothesis, H_0 : \mu = 23.9 years     {means that young women are not delaying marriage and marrying at a later age}

Alternate Hypothesis, H_A : \mu > 23.9 years     {means that young women are delaying marriage and marrying at a later age}

The test statistics that would be used here <u>One-sample z test statistics</u> as we know about the population standard deviation;

                         T.S. =  \frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }  ~ N(0,1)

where, \bar X = sample average age of the first time brides = 24.9 years

            \sigma = population standard deviation = 4.2 years

            n = sample of marriage records = 100

So, <u><em>the test statistics</em></u>  =  \frac{24.9-23.9}{\frac{4.2}{\sqrt{100} } }

                                     =  2.381

The value of z test statistics is 2.381.

<u>Now, at 1% significance level the z table gives critical value of and 2.326 for right-tailed test.</u>

Since our test statistic is more than the critical value of z as 2.381 > 2.326, so we have sufficient evidence to reject our null hypothesis as it will fall in the rejection region due to which <u>we reject our null hypothesis</u>.

Therefore, we conclude that young women are delaying marriage and marrying at a later age.

5 0
3 years ago
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