PLEASE HELP!!!!!!!!!!!!!!

Which statement is true?

marshall27 [118]

Answer:

B, the second answer choice

Step-by-step explanation:

Hope this helped! Please mark me as brainliest if you can!

4 0

3 years ago

I need some help with this! i want my grade to go up

mart [117]

Answer:

44.159

Step-by-step explanation:

7 0

2 years ago

Read 2 more answers

2. A coordinate grid is used to make a map of the park. The location of the drinking fountain is (-4, 2) ; the location of the b

leva [86]

Answer:

(A) The distance from the drinking fountain to the bench is 4, and the distance from the bench to the center of the slide is 3 because if you add -4 and 0 you will get 4 and if you subtract 5 and 2 you will get 3.

(B) The type of triangle that is created is an obtuse triangle because when you connect the 3 locations it creates an obtuse triangle on the coordinate grid.

Step-by-step explanation:

I already explained in the answer. Hope this helps and gets you an 100%

:)

5 0

2 years ago

Birth weights of babies born to full-term pregnancies follow roughly a Normal distribution. At Meadowbrook Hospital, the mean we

Marina86 [1]

Answer:

Required Probability = 0.1283 .

Step-by-step explanation:

We are given that at Meadow brook Hospital, the mean weight of babies born to full-term pregnancies is 7 lbs with a standard deviation of 14 oz.

Firstly, standard deviation in lbs = 14 ÷ 16 = 0.875 lbs.

Also, Birth weights of babies born to full-term pregnancies follow roughly a Normal distribution.

Let X = mean weight of the babies, so X ~ N( $\mu = 7 lbs , \sigma^{2} = 0.875^{2} lbs$ )

The standard normal z distribution is given by;

Z = $\frac{Xbar-\mu}{\frac{\sigma}{\sqrt{n} } }$ ~ N(0,1)

where, X bar = sample mean weight

n = sample size = 4

Now, probability that the average weight of the four babies will be more than 7.5 lbs = P(X bar > 7.5 lbs)

P(X bar > 7.5) = P( $\frac{Xbar-\mu}{\frac{\sigma}{\sqrt{n} } }$ > $\frac{7.5-7}{\frac{0.875}{\sqrt{4} } }$ ) = P(Z > 1.1428) = 0.1283 (using z% table)

Therefore, the probability that the average weight of the four babies will be more than 7.5 lbs is 0.1283 .

8 0

3 years ago

Of 585 samples of seafood purchased from various kinds of food stores in different regions of a country and genetically compared

Dafna1 [17]

Answer:

a) The 99% confidence interval is from 14.8% to 23.2%.

b) The confidence interval tells us that the true proportion of mislabeled is within 14.8% and 23.2%, with a 99% confidence. In other words, if we take samples of the same size, 99% of the samples will have a proportion within 0.148 and 0.232.

c) The confidence interval calculation take into account the sample size, so the width (or precision) of the interval depends on the sample size.

The only criticism that could be analyzed is to see if the sample is representative of the population.

Step-by-step explanation:

a) We have to calculate a 99% confidence interval for the proportion.

The sample proportion is p=0.19.

The standard error of the proportion is:

$\sigma_p=\sqrt{\dfrac{p(1-p)}{n}}=\sqrt{\dfrac{0.19*0.81}{585}}\\\\\\ \sigma_p=\sqrt{0.000263}=0.016$

The critical z-value for a 99% confidence interval is z=2.576.

The margin of error (MOE) can be calculated as:

$MOE=z\cdot \sigma_p=2.576 \cdot 0.02=0.042$

Then, the lower and upper bounds of the confidence interval are:

$LL=p-z \cdot \sigma_p = 0.19-0.042=0.148\\\\UL=p+z \cdot \sigma_p = 0.19+0.042=0.232$

The 99% confidence interval for the population proportion is (0.148, 0.232).

5 0

3 years ago