4x - 1 = 3 (x - 1) (help pls fast by tuesday!)

Find two numbers whose gcf is 90

Blababa [14]

That would be 90 and 180 i think

6 0

3 years ago

Read 2 more answers

If x represents a number, does 2/5 times x always represent 40% of that number?

marusya05 [52]

Answer:

Yes, it always represent 40% of that number.

Step-by-step explanation:

5 0

3 years ago

the marks in an examination for a Physics paper have normal distribution with mean μ and variance σ2 . 10% of the students obtai

Blababa [14]

Let $X$ be the random variable for the number of marks a given student receives on the exam.

10% of students obtain more than 75 marks, so

$P(X>75)=P\left(\dfrac{X-\mu}\sigma>\dfrac{75-\mu}\sigma\right)=P(Z>z_1)=0.10$

where $Z$ follows a standard normal distribution. The critical value for an upper-tail probability of 10% is

$P(Z>z_1)=1-F_Z(z_1)=0.10\implies z_1=F_Z^{-1}(0.90)$

where $F_Z(z)=P(Z\le z)$ denotes the CDF of $Z$ , and $F_Z^{-1}$ denotes the inverse CDF. We have

$z_1=F_Z^{-1}(0.90)\approx1.2816$

Similarly, because 20% of students obtain less than 40 marks, we have

$P(X$

so that

$P(Z$

Then $\mu,\sigma$ are such that

$\dfrac{75-\mu}\sigma\approx1.2816\implies75\approx\mu+1.2816\sigma$

$\dfrac{40-\mu}\sigma\approx-0.8416\implies40\approx\mu-0.8416\sigma$

and we find

$\mu\approx53.8739,\sigma\approx16.4848$

3 0

4 years ago

Help me please please

yanalaym [24]

Answer:

Hold on a sec i cant remember

3 0

3 years ago

In 1943, an organization surveyed 1100 adults and asked, "Are you a total abstainer from, or do you on occasion consume, alc

Fynjy0 [20]

Answer:

We conclude that the proportion of adults who totally abstain from alcohol has changed.

Step-by-step explanation:

We are given that in 1943, an organization surveyed 1100 adults and asked, "Are you a total abstainer from, or do you on occasion consume, alcoholic beverages?"

Of the 1100 adults surveyed, 429 indicated that they were total abstainers. In a recent survey, the same question was asked of 1100 adults and 352 in

Let p = proportion of adults who totally abstain from alcohol.

where, p = $\frac{429}{1100}$ = 0.39

So, Null Hypothesis, $H_0$ : p = 39% {means that the proportion of adults who totally abstain from alcohol has not changed}

Alternate Hypothesis, $H_A$ : p $\neq$ 39% {means that the proportion of adults who totally abstain from alcohol has changed}

The test statistics that would be used here 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 adults who totally abstain from alcohol = $\frac{352}{1100}$ = 0.32

n = sample of adults surveyed = 1100

So, test statistics = $\frac{0.32-0.39}{\sqrt{\frac{0.32(1-0.32)}{1100} } }$

= -4.976

The value of z test statistics is -4.976.

Now, at 0.10 significance level the z table gives critical values of -1.645 and 1.645 for two-tailed test.

Since our test statistics doesn't lie within the range of critical values of z, so we have sufficient evidence to reject our null hypothesis as it will fall in the rejection region due to which we reject our null hypothesis.

Therefore, we conclude that the proportion of adults who totally abstain from alcohol has changed.

4 0

3 years ago