All categories

2 years ago

Can someone please help and solve these 2 problems?

Mathematics

1 answer:

VladimirAG [237]2 years ago

8 0

Answer:

8 for 1

n is (0;8)

Step-by-step explanation:

look photo, but I'm not conviniat if n can be negative, if yes, i hope you will correct me

You might be interested in

Amelia rented a dvd and it was due to be returned on 26 november

SashulF [63]

Answer:

Step-by-step explanation:

I'm not familiar with the conversions from American dollars. I can help you solve it though.

Count how many days from Nov 26 to Dec 26 and multiply that by 9p.

It's 30 days by 9p. Take that total and convert it.

Hopefully this helps you.

8 0

2 years ago

Please help me with this surds question.

noname [10]

the answer is 0.2 or one fifths

3 0

2 years ago

Kona wants to bake 30 loaves of bananas bread and nut bread one is 2.50 and other is 2.75 she wants to make are lasts 44$ write

kykrilka [37]

Answer:$79.25

Step-by-step explanation:

6 0

2 years ago

Let F(x)=x^2-4 and G(x)=5-x Find (F/G)(x)=

sammy [17]

Answer:(F/G)(x)=7

Step-by-step explanation:

5 0

2 years ago

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