What are the answers?

Bev weighs a bag of apples labeled 5 pounds and finds that the weight is actually 72 ounces. To the nearest percent, what is the

Citrus2011 [14]

Answer: 10% is the percentage error.

Step-by-step explanation:

Since, According to the question the labeled weight of the a bag of apples = 5 pounds

Since, 1 pounds = 16 ounces

Therefore the labeled weight of the a bag of apples= 5 × 16 = 80 ounces

But, the actual weight of the bag of apples = 72 ounces

Since, The percentage error = ( experiment value - actual value)×100 /experiment value

Thus, percentage error of the weight of the bag of apple= (80-72)×100/80= 800/80= 10%

4 0

3 years ago

12a-4(5a-1) = 2(3a +6)-4a

Ipatiy [6.2K]

Answer:

The answet to the following question is a = -5/4.

You distribute and then combine similar terms.

7 0

3 years ago

Solve for m. 6 + 2 √5 6 - 2 √5 2 √5 - 6

lys-0071 [83]

M/2 - 5 = 3
M-2 /5= 6
M=6+2 /5
So the answer is the first equation

7 0

3 years ago

Weight gain during pregnancy. In 2004, the state of North Carolina released to the public a large data set containing informatio

pychu [463]

Answer:

1. B. H0: μ1−μ2=0, HA: μ1−μ2≠0

2. z=1.2114

3. P-value=0.2257

4. Do not reject H0

Step-by-step explanation:

We have to perfomr an hypothesis test to see if there is strong evidence that there is a significant difference between the two population means.

The null and alternative hypothesis are:

$H_0: \mu_1-\mu_2=0\\\\H_a: \mu_1-\mu_2\neq0$

Being μ1 the mean average gain for younger mothers and μ2 the mean average gain for mature mothers.

(NOTE: we are comparing means, not proportions, as it is the random variable is the weight gain).

As we are claiming "strong evidence", the level of significance will be 0.01.

For younger mothers, the sample size is n1=840, the sample mean is 30.7 and the sample standard deviation is s1=14.91.

For mature mothers, the sample size is n2=132, the sample mean is 29.15 and the sample standard deviation is s2=13.46.

The difference between means is

$M_d=\mu_1-\mu_2=30.7-29.15=1.55$

The standard error of the difference between means is

$s_M=\sqrt{\dfrac{s_1^2}{n_1}+\dfrac{s_2^2}{n_2}}=\sqrt{\dfrac{14.91^2}{840}+\dfrac{13.46^2}{132}}=\sqrt{ 0.2647+1.3725}=\sqrt{1.6372}\\\\\\s_M=1.2795$

Then, the statistic can be calculated as:

$z=\dfrac{M_d-(\mu_1-\mu_2)}{s_M}=\dfrac{1.55-0}{1.2795}=1.2114$

The P-value for this z-statistic in a two tailed test is:

$P-value=2P(z>1.2114)=0.2257$

As the P-value is greater than the significance level, the null hypothesis failed to be rejected.

There is no enough evidence to claim that the real average weight gain differs from mature and youger mothers.

5 0

4 years ago

Whats the parent and shift for y=x-6

Sonbull [250]

The parent function is either identity function or constant function
the translation is down 6

8 0

3 years ago