What is .74 repeating as a fraction?

Find the value of x :)))))

Nataly [62]

Answer:

A x =30

Step-by-step explanation:

(y+x) and (y-x) make a straight line so they add to 180 degrees

(y+x) + (y-x) = 180

Combine like terms

y+y+x-x=180

2y=180

Divide by 2

2y/2 = 180/2

y= 90

(y+x) and 2x make a straight line so they add to 180 degrees

y+x+2x = 180

Combine like terms

y+3x = 180

Since we know y we can substitute it in

90+3x = 180

Subtract 90 from each side

90-90+3x= 180-90

3x = 90

Divide by 3

3x/3 = 90/3

x= 30

5 0

3 years ago

As you move up a linear demand curve, the price elasticity of demand in absolute value A. decreases. B. increases. C. stays the

Andreyy89

Answer:

The answers to the questions are;

B. increases.

C. inelastic; increase.

Step-by-step explanation:

The price elasticity of demand for a linear demand in absolute value curve becomes smaller and smaller as we move downwards of the curve.

When the price elasticity of demand is calculated along a linear demand curve. This is so as for each pair of points at which the price elasticity of demand is calculated, the quantity demanded and the price change are somewhat similar, but as we move towards the top of the demand curve, the high prices and the low quantities shows the increase in demand elasticity.

Also in the inelastic region, as it can be shown by the areas of the rectangles formed by to adjacent price points, increase in price, increases the total revenue.

6 0

3 years ago

What is the anser for 194*33333

kkurt [141]

(the answer is) = 6466602

8 0

3 years ago

Rey feeds his dog 2/5 of a can of dog food in the morning and 4/5 of a can in the evening. How many cans of dog food will Rey ne

QveST [7]

Answer:

6 cans total if im correct

6 0

2 years ago

The mean MCAT score 29.5. Suppose that the Kaplan tutoring company obtains a sample of 40 students with a mean MCAT score of 32.

Paul [167]

Answer:

We conclude that the students that took the Kaplan tutoring have a mean score greater than 29.5.

Step-by-step explanation:

We are given that the Kaplan tutoring company obtains a sample of 40 students with a mean MCAT score of 32.2 with a standard deviation of 4.2.

Let $\mu$ = population mean score

So, Null Hypothesis, $H_0$ : $\mu \leq$ 29.5 {means that the students that took the Kaplan tutoring have a mean score less than or equal to 29.5}

Alternate Hypothesis, $H_A$ : $\mu$ > 29.5 {means that the students that took the Kaplan tutoring have a mean score greater than 29.5}

The test statistics that will be used here is One-sample t-test statistics because we don't know about population standard deviation;

T.S. = $\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }$ ~ $t_n_-_1$

where, $\bar X$ = sample mean MCAT score = 32.2

s = sample standard deviation = 4.2

n = sample of students = 40

So, the test statistics = $\frac{32.2-29.5}{\frac{4.2}{\sqrt{40} } }$ ~ $t_3_9$

= 4.066

The value of t-test statistics is 4.066.

Now, at 0.05 level of significance, the t table gives a critical value of 1.685 at 39 degrees of freedom for the right-tailed test.

Since the value of our test statistics is more than the critical value of t as 4.066 > 1.685, so we have sufficient evidence to reject our null hypothesis as it will fall in the rejection region.

Therefore, we conclude that the students that took the Kaplan tutoring have a mean score greater than 29.5.

3 0

3 years ago