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dem82 [27]
3 years ago
6

Translate the equation 6m = n to a verbal model.

Mathematics
2 answers:
klemol [59]3 years ago
7 0
6 multiplied by M equals the total. 
Katarina [22]3 years ago
7 0
Six m equals n hop this helps

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The scale factor of figure JKLMN to figure PQRST is 3:2. If KL = 9 cm and MN = 15 cm, what is the length of side ST?
Lina20 [59]
Since the scale factor is 3:2, and MN is 15, we can solve this.

The answer is 10.
7 0
4 years ago
A recipe for cookies calls for
tamaranim1 [39]
The answer is B.) you will need 3 cups of chocolate chips
7 0
3 years ago
A sector with a central angle measure of 7pi/4 (in radians) has a radius of 16 cm
strojnjashka [21]

Answer:

Area of sector = 704 cm ²

Step-by-step explanation:

Given:

Central angle = 7pi/4 (in radians)

Radius = 16 cm

Find:

Area of sector = ?

Computation:

Central angle = \frac{7}{4}\pi  \times \frac{180}{\pi } =315

Area\ of \ sector=\frac{\theta}{360}\pi  r^2\\\\Area\ of \ sector=\frac{315}{360}\pi  16^2\\\\Area\ of \ sector=704 cm^2

Area of sector = 704 cm ²

5 0
3 years ago
Mr thomas puts up fence posts from one end to another equal distances apart.there are 27 posts the width of each part is 10 cent
Fittoniya [83]
If there are 27 posts, then there are (n - 1) total spaces between the posts . . . 27 - 1 = 26 total spaces (since there has to be a post at each end of the line)

Here's an example of 7 posts equally spaced . . . so you can see there are one less spaces between posts than there are posts (i.e. 6 total spaces between 7 posts) . . .

[] . . . [] . . . [] . . . [] . . . [] . . . [] . . . []

27 posts = 27*(10cm) = 270 cm = 2.7m
26 spaces = 26*(30m) = 780 m

<u><em>Length of Fence = L = 780m + 2.7m = 782.7m</em></u>
6 0
4 years ago
Suppose the weights of apples are normally distributed with a mean of 85 grams and a standard deviation of 8 grams. The weights
user100 [1]

Answer:

a) 0.0304 = 3.04% probability a randomly chosen apple exceeds 100 g in weight.

b) The weight that 80% of the apples exceed is of 78.28g.

Step-by-step explanation:

Normal Probability Distribution:

Problems of normal distributions can be solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the z-score 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 p-value, we get the probability that the value of the measure is greater than X.

Weights of apples are normally distributed with a mean of 85 grams and a standard deviation of 8 grams.

This means that \mu = 85, \sigma = 8

a. Find the probability a randomly chosen apple exceeds 100 g in weight.

This is 1 subtracted by the p-value of Z when X = 100. So

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

Z = \frac{100 - 85}{8}

Z = 1.875

Z = 1.875 has a p-value of 0.9697

1 - 0.9696 = 0.0304

0.0304 = 3.04% probability a randomly chosen apple exceeds 100 g in weight.

b. What weight do 80% of the apples exceed?

This is the 100 - 80 = 20th percentile, which is X when Z has a p-value of 0.2, so X when Z = -0.84.

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

-0.84 = \frac{X- 85}{8}

X - 85 = -0.84*8

X = 78.28

The weight that 80% of the apples exceed is of 78.28g.

5 0
3 years ago
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