Ahahhajdockdndn help

Help someone plsss!!!!!!!!!!

mart [117]

Answer:

150 degrees.

Step-by-step explanation:

We know, 1 Hour angles = 15 Degrees

So, 10 hour walking = 10*15 = 150 degrees

Hope this will help. Please mark me brainliest.

8 0

3 years ago

In a random sample of high school seniors, the proportion who use text messaging was 0.88. In a random sample of high school fre

Bond [772]

Answer:

b. 0.02

Step-by-step explanation:

The smaller the p-value, the stronger the evidence that you should reject the null hypothesis. In this case, this will mean rejecting that the proportions are not significantly different.

Usually, a p-value is considered to be statistically significant when p ≤ 0.05.

From the answer options provided, alternative b. 0.02 is the only one that represents the difference in proportions to be statistically significant (there is only a 2% chance that the proportions are not significantly different).

Therefore, the answer is b. 0.02

5 0

3 years ago

The slope of a line is 5/9 and the slope of another line is -975. The two lines<br> are

Bond [772]

Answer:

the third option - they are perpendicular to each other.

Step-by-step explanation:

for a perpendicular slope we need to exchange the x and y values (remember, a slope is the ratio of y/x) and flip the sign.

and that is exactly what happened here.

3 0

3 years ago

A grocery store manager claims that 75% of shoppers purchase bananas as least once a month. Technology was used to simulate choo

Keith_Richards [23]

Answer:

P-value = 0.0333

At 5% level of significance;

0.0333 < 0.05

Therefore, we reject null hypothesis H₀ at 5% level of significance,

We conclude that proportion of shoppers who bought bananas at least once in the past month is overstated

Step-by-step explanation:

Given the data in the question;

To test whether population proportion p is overstated;

Null hypothesis H₀ : p = (75%) = 0.75

Alternative hypothesis H₁ : = < (75%) < 0.75

now, sample proportion p" = 64 / 100 = 0.64

from the dot plot below, we will determine the p-value for test { P(p" < 0.64)}

so, the number of times p"<0.64 in 150 simulations is 5

Hence; P(p" < 0.64 ) = 5 / 150 = 0.0333

P-value = 0.0333

At 5% level of significance;

0.0333 < 0.05

Therefore, we reject null hypothesis H₀ at 5% level of significance,

We conclude that proportion of shoppers who bought bananas at least once in the past month is overstated

7 0

3 years ago

A tablet PC contains 3217 music files. The distribution of file size is highly skewed with many small files. Suppose the true me

m_a_m_a [10]

Answer:

Let X the random variable who represents the file sizeof music. We know the following info:

$\mu =2.3,\sigma =3.25$

We select a sample of n=50 nails. That represent the sample size.

Since the sample size is large enough n >30, we can use the central limit theorem. From this theorem we know that the distribution for the sample mean $\bar X$ is given by:

$\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})$

And we can approximate the distribution of the sample mean as a normal distribution and no matter if the distribution for X is right skewed or no.

Step-by-step explanation:

Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The central limit theorem states that "if we have a population with mean μ and standard deviation σ and take sufficiently large random samples from the population with replacement, then the distribution of the sample means will be approximately normally distributed. This will hold true regardless of whether the source population is normal or skewed, provided the sample size is sufficiently large".

Let X the random variable who represents the file sizeof music. We know the following info:

$\mu =2.3,\sigma =3.25$

We select a sample of n=50 nails. That represent the sample size.

Since the sample size is large enough n >30, we can use the central limit theorem. From this theorem we know that the distribution for the sample mean $\bar X$ is given by:

$\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})$

And we can approximate the distribution of the sample mean as a normal distribution and no matter if the distribution for X is right skewed or no.

8 0

3 years ago