All categories

3 years ago

Pls can you solve this question from hegartymaths?

Mathematics

1 answer:

Sergeeva-Olga [200]3 years ago

8 0

The area of a circle with radius r is πr². So you need to find r such that

πr² = 56π

⇒ r² = 56

⇒ r = √56 = 2√14

You might be interested in

70 - q - q - 2q = 80???

Aneli [31]

70-q-q-2q= 80

combine like terms first

so you get 70-4q= 80

subtract 70 on both sides and you get -4q= 10

divide -4 on both sides and you get q= 10/-4 or -2.5

3 0

4 years ago

Is a 3.86 weighted gpa good for freshman year of highschool? I am in 4 honors classes, 1 AP, and 2 standard. Keep in mind this i

murzikaleks [220]

Answer:

barbie

Step-by-step explanation:

ken

4 0

3 years ago

Read 2 more answers

There are 104 apples in a bushel the probability of randomly choosing a granny smith apple from The Bushel is 25% how many apple

Sophie [7]

Answer:

There are 78 apples in the bushel that are not granny smith apple.

Step-by-step explanation:

We are given the following in the question:

Number of apples in the bushel,n = 104

Probability of a granny smith apple from The bushel = 25%

P(Granny smit apple) =

$p=0.25$

Thus, probability thta the apple is not a granny smith apple is

P( Not a granny smit apple) =

$q=1 - p = 1-0.25 = 0.75$

Number of apples that are not granny smit apple =

$=q\times n\\=0.75\times 104\\=78$

Thus, there are 78 apples in the bushel that are not granny smith apple.

5 0

3 years ago

Proxima Centauri, our nearest star except for the sun, is about 4·1013 km from Earth. Find to one decimal place its distance in

xenn [34]

Answer:

6.4 km hope that helps you bye bye

3 0

3 years ago

Read 2 more answers

In the United States, 35% of households own a 4K television. Suppose we take a random sample of 150 households. (a) Describe the

sattari [20]

Answer:

(a) The distribution of the sample proportion is Normal distribution.

(b) The probability that in this sample of 150 households that more than 50% own a 4K television is 0.00012.

Step-by-step explanation:

We are given that in the United States, 35% of households own a 4K television.

Suppose we take a random sample of 150 households.

Let $\hat p$ = sample proportion of households who own a 4K television.

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 households own a 4K television = 35%

n = sample of households = 150

(a) The distribution of the sample proportion is related to the Normal distribution.

(b) Probability that in this sample of 150 households more than 50% own a 4K television is given by = P( $\hat p$ > 0.50)

P( $\hat p$ > 0.50) = P( $\frac{ \hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ > $\frac{ 0.50-0.35}{\sqrt{\frac{0.50(1-0.50)}{150} } }$ ) = P(Z > 3.67) = 1 - P(Z $\leq$ 3.67)

= 1 - 0.99988 = 0.00012

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 = 3.67 in the z table which has an area of 0.99988.

Therefore, probability that in this sample of 150 households more than 50% own a 4K television is 0.00012.

8 0

3 years ago