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8_murik_8 [283]
3 years ago
13

Write 7/20 as a terminating decimal​

Mathematics
2 answers:
svetlana [45]3 years ago
7 0

Answer:

0.35

Step-by-step explanation:

storchak [24]3 years ago
4 0

Answer:

In order to do this, you would multiply

7

20

by

100

%

.

7

20

(

100

%

)

=

(

700

20

)

%

=

35

%

Multiplying

7

20

by

100

%

allows you to be able to use long division

→

you only multiply

100

by the numerator (the top number of the fraction).

To convert

7

20

to a decimal, simply divide

7

by

20

.

Step-by-step explanation:

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What adds to get 3 but multiplies to get negative 18
Colt1911 [192]
6 and -3

6 + -3 = 3

6  x -3 = -18

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X+16=24hvvcgcfcycdfdxfxxdgfv​
zaharov [31]

To do solve this you must isolate x. First subtract 16 to both sides (what you do on one side you must do to the other). Since 16 is being added to x, subtraction (the opposite of addition) will cancel it out (make it zero) from the left side and bring it over to the right side.

x + (16 - 16) = 24 - 16

x = 8

Check:

8 + 16 = 24

24 = 24

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7 0
3 years ago
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5 less than a number y is under 20.
klio [65]

Answer:

y - 5 < 20

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Step-by-step explanation:

y - 5 < 20

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4 0
4 years ago
Lexi is making an orange drink for party. She puts 4 pints of orange drink and 2 pints of water into a punch bowl. Her friend Ol
Nina [5.8K]
I think she can add 5p more of orange which is 9p and 6p more water which is 8p. I'm pretty sure this works because 9 is the next closest multiple of 3 and 8 is one number less than that. Hope this helps!
8 0
3 years ago
Navy PilotsThe US Navy requires that fighter pilots have heights between 62 inches and78 inches.(a) Find the percentage of women
Zigmanuir [339]

The first part of the question is missing and it says;

Use these parameters: Men's heights are normally distributed with mean 68.6 in. and standard deviation 2.8 in. Women's heights are normally distributed with mean 63.7 in. and standard deviation 2.9 in.

Answer:

A) Percentage of women meeting the height requirement = 72.24%

B) Percentage of men meeting the height requirement = 0.875%

C) Corresponding women's height =67.42 inches while corresponding men's height = 72.19 inches

Step-by-step explanation:

From the question,

For men;

Mean μ = 68.6 in

Standard deviation σ = 2.8 in

For women;

Mean μ = 63.7 in

Standard deviation σ = 2.9 in

Now let's calculate the standardized scores;

The formula is z = (x - μ)/σ

A) For women;

Z = (62 - 63.7)/2.9 = - 0.59

Z = (78 - 63.7)/2.9 = 4.93

The original question cam be framed as;

P(62 < X < 78).

So thus, the probability of only women will take the form of;

P(-0.59 < Z < 4.93) = P(Z<4.93) - P(Z > - 0.59)

From the normal probability table attached, when we interpolate, we'll arrive at P(Z<4.93) = 0.9999996

And P(Z > - 0.59) = 0.277595

Thus;

P(Z<4.93) - P(Z > - 0.59) =0.9999996 - 0.277595 = 0.7224

So, percentage of women meeting the height requirement is 72.24%.

B) For men;

Z = (62 - 68.6)/2.8 = -2.36

Z = (78 - 68.6)/2.8 = 3.36

Thus, the probability of only men will take the form of;

P(-2.36 < Z < 3.36) = P(Z<3.36) - P(Z > - 2.36)

From the normal probability table attached, when we interpolate, we'll arrive at P(Z<3.36) = 0.99961

And P(Z > -2.36) = 0.99086

Thus;

P(Z<3.36) - P(Z > -2.36) 0.99961 - 0.99086 = 0.00875

So, percentage of women meeting the height requirement is 72.24%.

B)For women;

Z = (62 - 63.7)/2.9 = - 0.59

Z = (78 - 63.7)/2.9 = 4.93

The original question cam be framed as;

P(62 < X < 78).

So thus, the probability of only women will take the form of;

P(-0.59 < Z < 4.93) = P(Z<4.93) - P(Z > - 0.59)

From the normal probability table attached, when we interpolate, we'll arrive at P(Z<4.93) = 0.9999996

And P(Z > - 0.59) = 0.277595

Thus;

P(Z<4.93) - P(Z > - 0.59) =0.9999996 - 0.277595 = 0.00875

So, percentage of women meeting the height requirement is 0.875%

C) Since the height requirements are changed to exclude the tallest 10% of men and the shortest10% of women.

For women;

Let's find the z-value with a right-tail of 10%. From the second table i attached ;

invNorm(0.90) = 1.2816

Thus, the corresponding women's height:: x = (1.2816 x 2.9) + 63.7= 67.42 inches

For men;

We have seen that,

invNorm(0.90) = 1.2816

Thus ;

Thus, the corresponding men's height:: x = (1.2816 x 2.8) + 68.6 = 72.19 inches

7 0
4 years ago
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