Joesph is 14 years younger than twice joesephens age. Joseph is 32 years old. how old is josephene?

Mathematics

2 answers:

soldi70 [24.7K]3 years ago

7 0

Answer:

Josephene is 23 years old

Step-by-step explanation:

First, you can assign the variables x and y to Joseph and Josephene, next you should write an equation. Since Joseph is 14 years younger than twice Josephene's age, you can write the equation x + 14 = 2y, if 32 is substituted in for x and the equation is simplified, you will end up with Josephene's age, 23.

I hope this helps :)

sashaice [31]3 years ago

4 0

Step-by-step explanation:

Joesphene is 23 years old, because: 2x=32+14=>2x=46=>x=23 .

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Anastasy [175]

The answer is 13!! :)

7 0

3 years ago

Read 2 more answers

The mean SAT score in mathematics, M, is 600. The standard deviation of these scores is 48. A special preparation course claims

kupik [55]

Answer:

Step-by-step explanation:

The mean SAT score is $\mu=600$ , we are going to call it \mu since it's the "true" mean

The standard deviation (we are going to call it $\sigma$ ) is

$\sigma=48$

Next they draw a random sample of n=70 students, and they got a mean score (denoted by $\bar x$ ) of $\bar x=613$

The test then boils down to the question if the score of 613 obtained by the students in the sample is statistically bigger that the "true" mean of 600.

- So the Null Hypothesis $H_0:\bar x \geq \mu$

- The alternative would be then the opposite $H_0:\bar x < \mu$

The test statistic for this type of test takes the form

$t=\frac{| \mu -\bar x |} {\sigma/\sqrt{n}}$

and this test statistic follows a normal distribution. This last part is quite important because it will tell us where to look for the critical value. The problem ask for a 0.05 significance level. Looking at the normal distribution table, the critical value that leaves .05% in the upper tail is 1.645.

With this we can then replace the values in the test statistic and compare it to the critical value of 1.645.

$t=\frac{| \mu -\bar x |} {\sigma/\sqrt{n}}\\\\= \frac{| 600-613 |}{48/\sqrt(70}}\\\\= \frac{| 13 |}{48/8.367}\\\\= \frac{| 13 |}{5.737}\\\\=2.266\\$

<h3>since 2.266>1.645 we can reject the null hypothesis.</h3>