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3 years ago

How many times greater is 9 x 105 than 8 x 102?

Mathematics

1 answer:

IgorLugansk [536]3 years ago

5 0

Answer:

Step-by-step explanation:

945 816 so 129 greater

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Please help it’s irritating when I keep having to post this and no ones bothering to even help me. Like I need to turn this is A

sukhopar [10]

Measure of E is congruent to K, so E is 50 degrees.

Measure of G is congruent to L, so G is 105 degrees.

Measure of F is congruent to J, J is 180 deg - 50 deg - 105 deg = 25 degrees, so F is 25 degrees.

The proof I am using is the Corresponding Angles Postulate.

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6 0

3 years ago

3 1/3 divided by 2/3 pls help me!!!! im stuck and pls show your work!

Vedmedyk [2.9K]

The answer would be 2 1/3 or 2.3 repeating or 7/3

8 0

3 years ago

Read 2 more answers

David has 36 apple and pear trees in his gardens. If he plant 4 more apple trees, he will have the same amount of apple and pear

amm1812

Answer:

16 apples, 20 pears

Step-by-step explanation:

36+4=40

So 40/2= 20 or an even amount of both apples and pears

20-4= 16 ~Original amount of apples

36-16=20 ~Amount of pears

6 0

3 years ago

Which of the following inequalities represents "the car has no more than 5 gallons of gas left in the gas tank"?

MrRissso [65]

The answer is B. Hope this helps :)

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3 years ago

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According to a study done by Wakefield Research, the proportion of Americans who can order a meal in a foreign language is 0.47.

UNO [17]

Answer:

Probability that the proportion of Americans who can order a meal in a foreign language is greater than 0.5 is 0.19766.

Step-by-step explanation:

We are given that according to a study done by Wake field Research, the proportion of Americans who can order a meal in a foreign language is 0.47.

Suppose a random sample of 200 Americans is asked to disclose whether they can order a meal in a foreign language.

Let $\hat p$ = sample proportion of Americans who can order a meal in a foreign language

The z-score probability distribution for sample proportion is given by;

Z = $\frac{ \hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ~ N(0,1)

where, $\hat p$ = sample proportion

p = population proportion of Americans who can order a meal in a foreign language = 0.47

n = sample of Americans = 200

Probability that the proportion of Americans who can order a meal in a foreign language is greater than 0.5 is given by = P( $\hat p$ > 0.50)

P( $\hat p$ > 0.06) = P( $\frac{ \hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ > $\frac{0.5-0.47}{\sqrt{\frac{0.5(1-0.5)}{200} } }$ ) = P(Z > 0.85) = 1 - P(Z $\leq$ 0.85)

= 1 - 0.80234 = 0.19766

Now, in the z table the P(Z $\leq$ x) or P(Z < x) is given. So, the above probability is calculated by looking at the value of x = 0.85 in the z table which has an area of 0.80234.

Therefore, probability that the proportion of Americans who can order a meal in a foreign language is greater than 0.5 is 0.19766.

4 0

3 years ago