Pretty please help!(:

the expression below gives the cost of green beans(x) and cucumbers (y). what is the cause of 6 bunches of green beans and 4 cuc

kipiarov [429]

This is my explanation thus sucks and half the timem you guys dont even have the right answer . So this is my answer for you at leatst try to get it tight because if you dont try you wont suceed

3 0

2 years ago

Britton rotated trapezoid ABCD 180° on the coordinate plane. If angle B is 20° and angle D is 90°, what is the degree measuremen

olasank [31]

Answer:

The answer is 20 degrees

Step-by-step explanation:

A trapezoid is a quadrilateral with four angles and the sum of the four angles is 360 degrees.

Britton rotated the trapezoid ABCD 180º, but we were not told that he tampered with the angles size or shape, we can now agree that it is still the same trapezoid that was rotated without any alterations in angles, size and shape.

Since we've been given that the angle B is 20º, and the measurement was not altered when rotating the trapezoid. Therefore the angle B' will also be 20º.

I hope this helps.

6 0

2 years ago

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jose is choosing a 3-letter password from the letters a,b,c,d. the password cannot have the same letter repeated in it. how many

kvasek [131]

Alright, so we are dealing with permutations. Permutations are the number of combinations in a specific order possible for the set.

There's 4 letters, each one can be paired with 3 other letters, but those letters could be in different orders. To figure out how many variations of each combination there are (aka the number of permutations) use this formula:

$_{n}P_{r}= \frac{n!}{(n-r)!}$

$r=$ number of elements in the subset

$n=$ number of elements in the set

$P=$ permutations of the set

There are only 3 elements in the subset because there is 1 that will not be repeated in each set, and there are 4 elements in the set.

Here's the math:

$_{n}P_{r}= \frac{n!}{(n-r)!}$

$_{4}P_{3}= \frac{4!}{(4-3)!}$

$_{4}P_{3}= \frac{4!}{(1)!}$

$_{4}P_{3}= \frac{4(3)(2)(1)}{1}$

$_{4}P_{3}=24$

There are 24 permutations. I can prove this by showing you the model:

ABCD, ABDC, ACBD, ACDB, ADBC, ADCB are the 6 arrangements possible of the set starting with the letter A. Because there are 4 letters, the total amount of permutations without repeated letters is 4 (letters) times 6 (possible arrangements), which equals 24.

Hope this helps!

Answer: 24 passwords are possible

5 0

2 years ago

"Immediately after a ban on using hand-held cell phones while driving was implemented, compliance with the law was measured. A r

sergiy2304 [10]

Answer:

(a) Null Hypothesis, $H_0$ : $p_1-p_2=0$ or $p_1= p_2$

Alternate Hypothesis, $H_A$ : $p_1-p_2\neq 0$ or $p_1\neq p_2$

(b) We conclude that there is a statistical difference in these two proportions measured initially and then one year later.

Step-by-step explanation:

We are given that a random sample of 1,250 drivers found that 98.9% were in compliance. A year after the implementation, compliance was again measured to see if compliance was the same (or not) as previously measured.

A different random sample of 1,100 drivers found 96.9% compliance."

Let $p_1$ = proportion of drivers that were in compliance initially

$p_2$ = proportion of drivers that were in compliance one year later

(a) Null Hypothesis, $H_0$ : $p_1-p_2=0$ or $p_1= p_2$ {means that there is not any statistical difference in these two proportions measured initially and then one year later}

Alternate Hypothesis, $H_A$ : $p_1-p_2\neq 0$ or $p_1\neq p_2$ {means that there is a statistical difference in these two proportions measured initially and then one year later}

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

T.S. = $\frac{(\hat p_1-\hat p_2)-(p_1-p_2)}{\sqrt{ \frac{\hat p_1(1-\hat p_1)}{n_1} + \frac{\hat p_2(1-\hat p_2)}{n_2}} }$ ~ N(0,1)

where, $\hat p_1$ = sample proportion of drivers in compliance initially = 98.9%

$\hat p_2$ = sample proportion of drivers in compliance one year later = 96.9%

$n_1$ = sample of drivers initially = 1,250

$n_2$ = sample of drivers one year later = 1,100

(b) So, the test statistics = $\frac{(0.989-0.969)-(0)}{\sqrt{ \frac{0.989(1-0.989)}{1,250} + \frac{0.969(1-0.969)}{1,100}} }$

= 3.33

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

P-value = P(Z > 3.33) = 1 - P(Z $\leq$ 3.33)

= 1 - 0.99957 = 0.00043

Since in the question we are not given with the level of significance so we assume it to be 5%. Now at 5% significance level, the z table gives critical values between -1.96 and 1.96 for two-tailed test.

Since our test statistics does not lies 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 there is a statistical difference in these two proportions measured initially and then one year later.

7 0

2 years ago

The photo shows a small coin. The scale from the actual coin to the photo is

Aleks [24]

8mm to 2cm is the same as 8mm to 20mm

8 : 20 simplifies to 2 : 5

3.25cm ÷ 5 = 0.65cm

0.65 x 2 = 1.3cm

Your answer is 1.3cm

6 0

2 years ago