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Advocard [28]
3 years ago
5

Pls need answer ASAP

Mathematics
1 answer:
FinnZ [79.3K]3 years ago
4 0

Answer:

8x +6    is the expression.....

when x=5

8(5) + 6 = 46

Step-by-step explanation:

hope this helps :3

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Alexander Graham Bell, the inventor of the telephone, also invented a kite made out of "cells' shaped like triangular pyramids.
IceJOKER [234]

Answer:

The fabric needed for each pyramid cell of the kite will have an area of 173 cm².

Step-by-step explanation:

Area of a face of the triangular pyramids = (1/2)(b)(h)

where

b = base of a triangle of the triangular pyramid = 20 cm

h = height of each triangle of the triangular pyramids = 17.3 cm

Area = (1/2)(20)(17.3) = 173 cm²

Just like the image presented with the question, each face of the pyramid that has a cell of fabric has only one, So, the fabric needed for each pyramid cell of the kite will have an area of 173 cm².

Hope this Helps!!!

5 0
3 years ago
Before she went on a diet, Jenna’s waist measured 32.45 inches. After the diet, it measured 26.97 inches. How many inches did Je
mash [69]

Answer:

5.48

Step-by-step explanation:

big number - little number

6 0
4 years ago
Read 2 more answers
Write an equation in slope-intercept form of the line that passes through the given points.
tensa zangetsu [6.8K]

Answer:

y = 1/3x -3

Step-by-step explanation:

3 0
3 years ago
Help fast thanks In the figure, prove that m∥n
astraxan [27]

Answer:

See below

Step-by-step explanation:

a + 133 = 180 because they are supplementary angles. (adjacent angles that form a straight angle)

a = 47 (you substract 133 at each side of the previous equation, leaving that a = 47)

m || n Since "a" measure the same as the angle in the figure that measures 47 both are corresponding angles, therefore m and n are parallels

7 0
3 years ago
Read 2 more answers
The lengths of pregnancies are normally distributed with a mean of 266 days and a standard deviation of 15 days.
inessss [21]

The lengths of pregnancies are normally distributed with a mean of 266 days and a standard deviation of 15 days.

That is,

Consider X be the length of the pregnancy

Mean and standard deviation of the length of the pregnancy.

Mean \mu =266\\

Standard deviation \sigma =15

For part (a) , to find the probability of a pregnancy lasting 308 days or longer:

That is, to find P(X\geq 308)

Using normal distribution,

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

z=\frac{308-266}{15}

=\frac{42}{15}

Thus z=2.8

So P\left (X\geq 308  \right )=1-P(X

=1-P(z

=1-Table\:  value\:  of\:  2.8

=1-0.99744

=0.00256

Thus the probability of a pregnancy lasting 308 days or longer is given by 0.00256.

This the answer for part(a): 0.00256

For part(b), to find the length that separates premature babies from those who are not premature.

Given that the length of pregnancy is in the lowest 3​%.

The z-value for the lowest of 3% is -1.8808

Then X=\frac{X-\mu}{\sigma}\Rightarrow X=z*\sigma+\mu

This implies X=-1.8808*15+266=237.788

Thus the babies who are born on or before 238 days are considered to be premature.

5 0
3 years ago
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