All categories

3 years ago

A number that's greater than 40 whose prime factorisation contains 3 prime number that don't repeat

Mathematics

1 answer:

katovenus [111]3 years ago

3 0

42. Because you divide into 6 and 7. 6 is divided into 3 and 2. So, 2 times 3 times 7.

You might be interested in

2-5=x/1.2 (please solve this)

Leni [432]

Step-by-step explanation:

x=3.6

semoga bermanfaat

meet belajar skuy

6 0

3 years ago

Andrew is selling tickets for the school play, and charges t dollars for adult tickets and 2 dollars for child tickets. On Monda

Andru [333]

Answer:

T=4 plz make brainlest answer

Step-by-step explanation:

Monday overall cost 36+5t

Tuesday overall cost 20+9t

36+5t=20+9t

-20 -20

16+5t =9t

-5t -5t

16=4t

t=4

Check

36+5t=20+9t (t=4)

36+20=20+36

It works

3 0

3 years ago

What's is the converse of the statement?If x is odd, then 2x is even.

dsp73

If you have two statements p and q and $p\Rightarrow q$ is true, then $\neg q\Rightarrow \neg p$ is also true.

In your case, statements are p - "x is odd" and q - "2x is even".

Then $\neg p$ - "x is not odd" and $\neg q$ - "2x is not even."

Hence $\neg q\Rightarrow \neg p$ sounds as: "If 2x is not even, then x is not odd".
Answer: correct choice is D.

7 0

3 years ago

After a college football team once again lost a game to their archrival, the alumni association conducted a survey to see if alu

valentina_108 [34]

Answer:

P-value for this hypothesis test is 0.00175.

Step-by-step explanation:

We are given that the alumni association conducted a survey to see if alumni were in favor of firing the coach.

A simple random sample of 100 alumni from the population of all living alumni was taken. Sixty-four of the alumni in the sample were in favor of firing the coach.

Let p = proportion of all living alumni who favored firing the coach

SO, Null Hypothesis, $H_0$ : p = 0.50 {means that the majority of alumni are not in favor of firing the coach}

Alternate Hypothesis, $H_A$ : p > 0.50 {means that the majority of alumni are in favor of firing the coach}

The test statistics that will be used here is One-sample z proportion statistics;

T.S. = $\frac{\hat p-p}{{\sqrt{\frac{\hat p(1-\hat p)}{n} } } } }$ ~ N(0,1)

where, $\hat p$ = sample proportion of the alumni in the sample who were in favor of firing the coach = $\frac{64}{100}$ = 0.64

n = sample of alumni = 100

So, test statistics = $\frac{0.64-0.50}{{\sqrt{\frac{0.64(1-0.64)}{100} } } } }$

= 2.92

Now, P-value of the hypothesis test is given by ;

P-value = P(Z > 2.92) = 1 - P(Z $\leq$ 2.92)

= 1 - 0.99825 = 0.00175

Therefore, the P-value for this hypothesis test is 0.00175.

4 0

3 years ago

The time to complete a standardized exam in the BYU-Idaho Testing Center is approximately normal with a mean of 70 minutes and a

leva [86]

Answer:

Step-by-step explanation:

Given that X the time to complete a standardized exam in the BYU-Idaho Testing Center is approximately normal with a mean of 70 minutes and a standard deviation of 10 minutes.

We have 68 rule as 2/3 of total would lie within 1 std deviation, and 95 rule as nearly 95% lie within 2 std deviations from the mean.

We have std deviation = 10

Hence 2 std deviations from the mean

= Mean ±2 std deviations

= $70$ ±20

= $(50,90)$

Below 50, 0.25 or 2.5% would complete the exam.

3 0

3 years ago