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3 years ago

Look at the picture and thanks

Mathematics

2 answers:

disa [49]3 years ago

6 0

Answer:

Step-by-step explanation:

Sjsjdjddjdjdjjddjdjdjdjdjdndnsnsnsbs answer is E

yawa3891 [41]3 years ago

4 0

Answer: C i think

Step-by-step explanation: hop this helps :)

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What is (radical 75 y 19 ) in the simplest form.

Liono4ka [1.6K]

Answer:

7519 is already in the simplest form. It can be written as 3.947368 in decimal form (rounded to 6 decimal places).

Step-by-step explanation:

3 0

3 years ago

when 200 apples are sold at Rs 1.50 each there will be rupees hundred loss how many apples should be sold for rupees 150 to gain

Lera25 [3.4K]

Answer:

Step-by-step explanation:

Sales price of 200 apples = 300

At this price, there was loss of Rs. 100

Means the cost of the apples was

300 + 100 = Rs. 400

To earn Rs. 100, they should have been sold for Rs. 500

4 0

3 years ago

Simplify the expression: -2(2c-4)-9c - 12

Fantom [35]

Answer:

-13c-4

Step-by-step explanation:

8 0

3 years ago

Which point satisfies the system of equations y = 3x − 2 and y = -2x + 3?

Elodia [21]

Meet of lines: y = 3x − 2 and y = -2x + 3?

y=3x-2 has slope 3 and y intercept -2 (which means thru (0,-2)) , so that's the blue line through point D.

y=-3x+3 has slope -2 and y intercept 3, so the red line through points A, C and D.

The point D(1,1) is the intersection of the lines. Let's check that algebraically:

3x-2=3(1)-2=1=y, good

-2x+3 = -2(1)+3=1=y, good

Answer: D

8 0

3 years ago

Suppose a simple random sample of size nequals36 is obtained from a population with mu equals 74 and sigma equals 6. (a) Descri

OlgaM077 [116]

Part a)

The simple random sample of size n=36 is obtained from a population with

$\mu = 74$

and

$\sigma = 6$

The sampling distribution of the sample means has a mean that is equal to mean of the population the sample has been drawn from.

Therefore the sampling distribution has a mean of

$\mu = 74$

The standard error of the means becomes the standard deviation of the sampling distribution.

$\sigma_ { \bar X } = \frac{ \sigma}{ \sqrt{n} } \\ \sigma_ { \bar X } = \frac{ 6}{ \sqrt{36} } = 1$

Part b) We want to find

$P(\bar X \:>\:75.9)$

We need to convert to z-score.

$P(\bar X \:>\:75.9) = P(z \:>\: \frac{75.9 - 74}{1} ) \\ = P(z \:>\: \frac{75.9 - 74}{1} ) \\ = P(z \:>\: 1.9) \\ = 0.0287$

Part c)

We want to find

$P(\bar X \: < \:71.95)$

We convert to z-score and use the normal distribution table to find the corresponding area.

$P(\bar X \: < \:71.95) = P(z \: < \: \frac{71.9 5- 74}{1} ) \\ = P(z \: < \: \frac{71.9 5- 74}{1} ) \\ = P(z \: < \: - 2.05) \\ = 0.0202$

Part d)

We want to find :

$P(73\:$

We convert to z-scores and again use the standard normal distribution table.

$P( \frac{73 - 74}{1} \:< \: z$

5 0

3 years ago