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3 years ago

Helpppppp I’ll Mark you brainliest

Mathematics

2 answers:

mixer [17]3 years ago

5 0

Answer:

6.32

Step-by-step explanation:

I counted up from G to H and I got 6. Since the line is a diagonal it makes it odd as in a weird number like a decimal most times. It can't be 6 because the line wouldn't make it to H it would be just short. 5.83 is already too short. 8 is to long for the line to be as it is now as a diagonal. 8, you would have to count up 6 and 2 to the right perfect but the line doesn't do that.

Debora [2.8K]3 years ago

3 0

Answer:

6.32

Step-by-step explanation:

You might be interested in

A right circular cylinder has a volume of 500 cu in. If the base has a radius of 4 in., what's the altitude of the cylinder? Rou

GarryVolchara [31]

Cylinder Volume = π • r² • height
height = Cylinder Volume / (PI * radius^2)
height = 500 cubic inches / 3.14159 * 16
height = 9.9471839432 
height = 9.95 (rounded)

4 0

3 years ago

Please help! I need answers for both questions it’s very confusing!

Lilit [14]

Answer:

for number 5 its:

-15 -8 8/9 -3.3

and for number 6 its:

-3.3 -8 8/9 and -15

also sorry if its wrong

5 0

2 years ago

Read 2 more answers

Rockwell hardness of pins of a certain type is known to have a mean value of 50 and a standard deviation of 1.2.a. If the distri

zalisa [80]

Answer:

$P(\= X \ge 51 ) =0.0062$

$P(\= X \ge 51 ) = 0$

Step-by-step explanation:

From the question we are told that

The mean value is $\mu = 50$

The standard deviation is $\sigma = 1.2$

Considering question a

The sample size is n = 9

Generally the standard error of the mean is mathematically represented as

$\sigma_x = \frac{\sigma }{\sqrt{n} }$

=> $\sigma_x = \frac{ 1.2 }{\sqrt{9} }$

=> $\sigma_x = 0.4$

Generally the probability that the sample mean hardness for a random sample of 9 pins is at least 51 is mathematically represented as

$P(\= X \ge 51 ) = P( \frac{\= X - \mu }{\sigma_{x}} \ge \frac{51 - 50 }{0.4 } )$

$\frac{\= X -\mu}{\sigma } = Z (The \ standardized \ value\ of \ \= X )$

$P(\= X \ge 51 ) = P( Z \ge 2.5 )$

=> $P(\= X \ge 51 ) =1- P( Z < 2.5 )$

From the z table the area under the normal curve to the left corresponding to 2.5 is

$P( Z < 2.5 ) = 0.99379$

=> $P(\= X \ge 51 ) =1-0.99379$

=> $P(\= X \ge 51 ) =0.0062$

Considering question b

The sample size is n = 40

Generally the standard error of the mean is mathematically represented as

$\sigma_x = \frac{\sigma }{\sqrt{n} }$

=> $\sigma_x = \frac{ 1.2 }{\sqrt{40} }$

=> $\sigma_x = 0.1897$

Generally the (approximate) probability that the sample mean hardness for a random sample of 40 pins is at least 51 is mathematically represented as

$P(\= X \ge 51 ) = P( \frac{\= X - \mu }{\sigma_x} \ge \frac{51 - 50 }{0.1897 } )$

=> $P(\= X \ge 51 ) = P(Z \ge 5.2715 )$

=> $P(\= X \ge 51 ) = 1- P(Z < 5.2715 )$

From the z table the area under the normal curve to the left corresponding to 5.2715 and

=> $P(Z < 5.2715 ) = 1$

$P(\= X \ge 51 ) = 1- 1$

=> $P(\= X \ge 51 ) = 0$

5 0

3 years ago

Vector component of ū and w are as

Helen [10]

Answer:

r = ( -14.4 , 6.47 )

Step-by-step explanation:

in the photo you will find all the steps

7 0

2 years ago

AlMarai, a beverage company is planning to change its juice package size. Currently, there are 9 ounces of juice in each package

Andre45 [30]

The percentage of more juices that will be in the new package is 35.71%

<h3>What are word problems?</h3>

Word problems in mathematics are things happening in our daily life and we can use mathematical concepts and arithmetic operations to put a solution to those things.

From the given information, it was stated that:

Almarai usually have 9 ounces of juice in each package, while
Their new package will contain 14 ounces.

Our objective is to find the percentage of more juice that will be in the new package.

The percentage of more juice in the new package will be:

$\mathbf{=\dfrac{14 - 9}{14} \times 100 \%}$

$\mathbf{=\dfrac{5}{14} \times 100 \%}$

= 35.71%

Learn more about word problems here:

brainly.com/question/21405634

7 0

2 years ago