All categories

3 years ago

At a zoo, 7 lizards ate 35 grams of insects in one

Mathematics

1 answer:

lapo4ka [179]3 years ago

6 0

Answer 180 grams of insects

Explanation 7 •\• by 35 = 5 grams each/night.
36 x 5= 180 grams

You might be interested in

Which equation has two real solutions?

Otrada [13]

C and d is indubitably correct

5 0

3 years ago

Hiii please help i’ll give brainliest if you give a correct answer please thanks!

yKpoI14uk [10]

Answer:

<h3>Step-by-step explanation: for #2, look at 6 3/4 and make it into a decimal. Do the same to 7 1/2. Compare them both to see which one is shorter by subtracting 6 3/4 from 7 1/2 and when you get your answer for that, make that into a fraction. Predict that 6 3/4 is shorter because of how much less it is compared to 7 1/2. (I hope I helped you in some way. Good luck!)</h3>

7 0

3 years ago

Shelly biked 21 miles in 4 hours (part A) what is Shelly's average speed in miles per hour (part B) at the same rate how many ho

Alex73 [517]

A) Divide total miles by total time:

21 miles / 4 hours = 5.25 miles per hour.

B) Divide miles by miles per hour:

42 miles / 5.25 miles per hour = 8 hours.

3 0

3 years ago

A simple random sample of 20 third-grade children from a certain school district is selected, and each is given a test to measur

ICE Princess25 [194]

Answer:

Option C) 5.62 points

Step-by-step explanation:

We are given the following in the question:

Sample size, n = 20

Sample mean score = 64

Sample standard deviation, s = 12

Degree of freedom =

$=n-1\\=20-1\\=19$

We have to calculate margin of error for a 95% confidence interval.

Formula for margin of error:

$t_{critical}\times \displaystyle\frac{s}{\sqrt{n}}$

Putting the values, we get,

$t_{critical}\text{ at degree of freedom 19 and}~\alpha_{0.05} = \pm 2.093$

$2.093\times (\displaystyle\frac{12}{\sqrt{20}} )\\\\ =5.6161\approx 5.62$

Thus, the correct answer is

Option C) 5.62 points

4 0

3 years ago

The service life of a battery used in a cardiac pacemaker is assumed to be normally distributed. A random sample of ten batterie

Fittoniya [83]

Answer:

The confidence interval is $25.16 < \mu < 26.85$

Step-by-step explanation:

From the question we are given a data set

25.5, 26.1, 26.8, 23.2, 24.2, 28.4, 25.0, 27.8, 27.3, and 25.7.

The mean of the this sample data is

$\= x = \frac{\sum x_i}{n}$

where is the sample size with values n = 10

$\= x = \frac{25.5+ 26.1+ 26.8+23.2+ 24.2+ 28.4+ 25.0+ 27.8+ 27.3+ 25.7}{10}$

$\= x = 26$

The standard deviation is evaluated as

$\sigma = \sqrt{\frac{\sum (x-\= x)}{n} }$

substituting values

$= \sqrt{\frac{ ( 25.5-26)^2, (26.1-26)^2, (26.8-26)^2, (23.2-26)^2}{10} }$

$\cdot \ \cdot \ \cdot \sqrt{\frac{ ( 24.2-26)^2, (28.4-26)^2+( 25.0-26)^2+ (27.8-26)^2+( 27.3-26)^2+( 25.7-26)^2}{10} }$

$\sigma = 1.625$

The confidence level is given as 90% hence the level of significance is calculated as

$\alpha = 100 -90$

$\alpha =10$ %

$\alpha = 0.10$

Now the critical values of $\frac{\alpha }{2}$ is obtained from the normal distribution table as

$Z_{\frac{\alpha }{2} } = 1.645$

The reason we are obtaining the critical values of $\frac{\alpha }{2}$ instead of $\alpha$ is because we are considering two tails of the area under the normal curve