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3 years ago

Please I need help badly and quickly, pls do not answer if you just want the points. I need an accurate answer for this. Cheers

Mathematics

1 answer:

natita [175]3 years ago

7 0

Answer

Q11 is 50,3

Q12, 23,19

Step-by-step explanation:

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Pythagorean theorem

Talja [164]

Answer:

b=20 yards

Step-by-step explanation:

15^2+x^2=25^2

225+x^2=625

x^2=400

x=20

5 0

3 years ago

The dot plots below show the weights of the players of two teams:

Nadusha1986 [10]

Answer:B

Step-by-step explanation:

3 0

3 years ago

Read 2 more answers

Sweets are sold loose, or pre-packed in 120g bags. The 120g bags are ?1.49 each. The loose sweets are ?0.89 for 100g. By calcula

tester [92]

Answer:

The loose sweets at ?0.89 for 100 g.

Step-by-step explanation:

First, calculate the price per gram. You do this by dividing the price by the grams.

?1.49 / 120 g = 1.49 / 120 = 0.0124 (4 dp)

Because the answer was very long, I have rounded it to 4 decimal places (4 dp).

?0.89 / 100 g = 0.89 / 100 = 0.0089

Next, you must calculate both pre-packed and loose sweets to the same weight. I am calculating them both to 100 g.

0.0124 x 100 = 1.24

0.0089 x 100 = 0.89

Finally, the cheapest product for 100 g will be the better value. In this case, it is the loose sweets.

6 0

3 years ago

Use the Pythagorean theorem to find c!!!!

Troyanec [42]

A^2 + b^2 = c^2
6^2 + 8^2 = c^2
36 + 8^2 = c^2
36 + 64 = c^2
100 = c^2
√100 = c
10 = c

hope this helped, pls mark brainliest :)

6 0

3 years ago

A company produces steel rods. The lengths of the steel rods are normally distributed with a mean of 201.9-cm and a standard dev

Nesterboy [21]

Answer:

There is a 0.08% probability that the average length of a randomly selected bundle of steel rods is greater than 204.1-cm.

Step-by-step explanation:

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

A company produces steel rods. The lengths of the steel rods are normally distributed with a mean of 201.9-cm and a standard deviation of 2.1-cm. This means that $\mu = 201.9, \sigma = 2.1$ .

For shipment, 9 steel rods are bundled together. Find the probability that the average length of a randomly selected bundle of steel rods is greater than 204.1-cm.

By the Central Limit Theorem, since we are using the mean of the sample, we have to use the standard deviation of the sample in the Z formula. That is:

$s = \frac{\sigma}{\sqrt{n}} = \frac{2.1}{\sqrt{9}} = 0.7$

This probability is 1 subtracted by the pvalue of Z when $X = 204.1$ .

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{204.1 - 201.9}{0.7}$

$Z = 3.14$

$Z = 3.14$ has a pvalue of 0.9992. This means that there is a 1-0.9992 = 0.0008 = 0.08% probability that the average length of a randomly selected bundle of steel rods is greater than 204.1-cm.

3 0

3 years ago