Help im doing a math test

Steve changed 600 US dollars into 480 Euros. How many Euros did he receive for each US dollar?

Vlad [161]

Answer:

the answer is 4/5 which is .8

Step-by-step explanation:

6 0

4 years ago

Read 2 more answers

Do political science classes require less writing than history classes? The 55 randomly selected political science classes assig

saul85 [17]

Answer:

We conclude that political science classes require less writing time than history classes at 0.01 level of significance.

Step-by-step explanation:

We are given that the 55 randomly selected political science classes assigned an average of 16.3 pages of essay writing for the course. The standard deviation for these 55 classes was 4.8 pages.

The 54 randomly selected history classes assigned an average of 18.9 pages of essay writing for the course. The standard deviation for these 54 classes was 5 pages.

Let $\mu_1$ = average writing time required by political science classes.

$\mu_2$ = average writing time required by history classes.

So, Null Hypothesis, $H_0$ : $\mu_1\geq\mu_2$ {means that political science classes require more or equal writing time than history classes}

Alternate Hypothesis, $H_A$ : $\mu_1$ {means that political science classes require less writing time than history classes}

The test statistics that would be used here Two-sample t test statistics as we don't know about the population standard deviation;

T.S. = $\frac{(\bar X_1-\bar X_2)-(\mu_1-\mu_2)}{s_p \sqrt{\frac{1}{n_1}+\frac{1}{n_2} } }$ ~ $t__n_1_-_n_2_-_2$

where, $\bar X_1$ = sample average pages of essay writing for political science classes = 16.3 pages

$\bar X_2$ = sample average pages of essay writing for history classes = 18.9 pages

$s_1$ = sample standard deviation for political science classes = 4.8 pages

$s_2$ = sample standard deviation for history classes = 5 pages

$n_1$ = sample of political science classes = 55

$n_2$ = sample of history classes = 54

Also, $s_p=\sqrt{\frac{(n_1-1)s_1^{2}+(n_2-1)s_2^{2} }{n_1+n_2-2} }$ = $\sqrt{\frac{(55-1)\times 4.8^{2}+(54-1)\times 5^{2} }{55+54-2} }$ = 4.90

So, test statistics = $\frac{(16.3-18.9)-(0)}{4.9 \sqrt{\frac{1}{55}+\frac{1}{54} } }$ ~ $t_1_0_7$

= -2.77

The value of t test statistics is -2.77.

Now, at 0.01 significance level the t table gives critical value of -2.365 at 107 degree of freedom for left-tailed test.

Since our test statistics is less than the critical value of t, 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 political science classes require less writing time than history classes.

3 0

3 years ago

ASAP WHO EVER ANSWERS THIS WILL RECIEVE BRANLIEST

Masteriza [31]

The answer is 50
The ratio of boys to girls is 10:25 in the school
If sample has 20 boys (10x2) then based on ratio it would likely have 50 girls (25x2)

5 0

3 years ago

An Internet reaction time test asks subjects to click their mouse button as soon as a light flashes on the screen. The light is

Ganezh [65]

Answer:

(a) 0.20

(b) 31%

(c) 2.52 seconds

Step-by-step explanation:

The random variable Y models the amount of time the subject has to wait for the light to flash.

The density curve represents that of an Uniform distribution with parameters a = 1 and b = 5.

So, $Y\sim Unif(1,5)$

(a)

The area under the density curve is always 1.

The length is 5 units.

Compute the height as follows:

$\text{Area under the density curve}=\text{length}\times \text{height}$

$1=5\times\text{height}\\\\\text{height}=\frac{1}{5}\\\\\text{height}=0.20$

Thus, the height of the density curve is 0.20.

(b)

Compute the value of P (Y > 3.75) as follows:

$P(Y>3.75)=\int\limits^{5}_{3.75} {\frac{1}{b-a}} \, dy \\\\=\int\limits^{5}_{3.75} {\frac{1}{5-1}} \, dy\\\\=\frac{1}{4}\times [y]^{5}_{3.75}\\\\=\frac{5-3.75}{4}\\\\=0.3125\\\\\approx 0.31$