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Kamila [148]
3 years ago
7

What two consecutive whole numbers lie between the square root of 55?

Mathematics
2 answers:
Kamila [148]3 years ago
6 0
\boxed{7 \ \textless \  \sqrt{55} \ \textless \  8}\\\\because\ 7^2=49 \ \textless \  55\ and\ 8^2=64 \ \textgreater \  55
Orlov [11]3 years ago
6 0
By using directly calculators that are capable of solving for square roots, we know that the value of the square root of 55 is approximately equal to 7.42. The whole numbers that are directly to the left and right of 7.42 in a number line are 7 and 8. Thus, the answer is 7 and 8. 
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Todd uses 21 white tiles & some black tiles to make a mosaic.The mosaic has a total area of 144 square centimeters.Each tile
icang [17]

Answer: Todd used 27 black tiles.

Step-by-step explanation:

Let x represent the number of black tiles that Todd used.

Todd uses 21 white tiles & some black tiles to make a mosaic. This means that the total number of white and black tiles that Todd used is 21 + x

Each tile has an area of 3 square centimeters. This means that the total area covered by the white tiles is

21 × 3 = 63 square centimeters

Also, the total area covered by the black tiles is

3 × x = 3x square centimeters

The mosaic has a total area of 144 square centimeters. This means that

3x + 63 = 144

3x = 144 - 63

3x = 81

x = 81/3 = 27

3 0
3 years ago
write the ratio as a fraction in simplest form, with whole numbers in the numerator and denominator. 40 to 56
masha68 [24]

Answer:

  5/7

Step-by-step explanation:

A common factor of 8 can be canceled from numerator and denominator.

  40/56 = (5·8)/(7·8) = (5/7)·(8/8) = (5/7)·1 = 5/7

_____

Since you know your multiplication tables, you know that 40 and 56 are both multiples of 8.

__

If you don't know your multiplication tables, you can find the greatest common divisor (GCD) of the two numbers and divide each by that. The GCD can be found using Euclid's algorithm. For that, you divide the larger by the smaller and use the remainder as the new smaller number. The original smaller number is now the larger number. For these numbers, that looks like ...

  56 ÷ 40 = 1 r 16

  40 ÷ 16 = 2 r 8

  16 ÷ 8 = 2 r 0 . . . . . the zero remainder signals that the divisor (8) is the GCD

Now, your fraction is ...

  (40/8) / (56/8) = 5/7

3 0
3 years ago
Read 2 more answers
Please help me answer the question 9 though 10
olchik [2.2K]

Answer:

9) slope of AC = slope of FH

Equation of the line is y = - x + 5

10) Area = 320

Step-by-step explanation:

For 9, the only thing similar IS the slope of the hypotenuse because the triangles are along the same line. Unless they're asking for the equation of the line, then that's just using the point slope equation which is:

y = - x + 5

For 10, just multiply 4 and 5 by 4 to get 16 and 20. Then multiply 16 and 20 to get 320.

8 0
3 years ago
A practice law exam has 100 questions, each with 5 possible choices. A student took the exam and received 13 out of 100.If the s
Cloud [144]

Answer:

z=\frac{13-20}{4}=-1.75

Assuming:

H0: \mu \geq 20

H1: \mu

p_v = P(Z

Step-by-step explanation:

The binomial distribution is a "DISCRETE probability distribution that summarizes the probability that a value will take one of two independent values under a given set of parameters. The assumptions for the binomial distribution are that there is only one outcome for each trial, each trial has the same probability of success, and each trial is mutually exclusive, or independent of each other".

Let X the random variable of interest (number of correct answers in the test), on this case we now that:

X \sim Binom(n=100, p=0.2)

The probability mass function for the Binomial distribution is given as:

P(X)=(nCx)(p)^x (1-p)^{n-x}

Where (nCx) means combinatory and it's given by this formula:

nCx=\frac{n!}{(n-x)! x!}

We need to check the conditions in order to use the normal approximation.

np=100*0.2=20 \geq 10

n(1-p)=20*(1-0.2)=16 \geq 10

So we see that we satisfy the conditions and then we can apply the approximation.

If we appply the approximation the new mean and standard deviation are:

E(X)=np=100*0.2=20

\sigma=\sqrt{np(1-p)}=\sqrt{100*0.2(1-0.2)}=4

So we can approximate the random variable X like this:

X\sim N(\mu =20, \sigma=4)

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".  The letter \phi(b) is used to denote the cumulative area for a b quantile on the normal standard distribution, or in other words: \phi(b)=P(z

The z score is given by this formula:

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

If we replace we got:

z=\frac{13-20}{4}=-1.75

Let's assume that we conduct the following test:

H0: \mu \geq 20

H1: \mu

We want to check is the score for the student is significantly less than the expected value using random guessing.

So on this case since we have the statistic we can calculate the p value on this way:

p_v = P(Z

5 0
3 years ago
How do I solve this?
maksim [4K]

Answer:

Answer A represents this

x <= -3  ;  x > 9

Step-by-step explanation:

Isolate x on each of the inequalities:

4 x + 4 <= - 8

subtract 4 from both sides

4 x <= - 12

divide both sides by 4 to isolate x completely (notice 4 is positive, so there is NO flipping of the inequality symbol)

x <= -3

This inequality is represented by shading the left section of the number line starting at x = -3 (make sure you draw a SOLID dot to mae clear that the point x = -3 is also included in your drawing.

The other inequality:

11 x - 11 > 88

add 11 to both sides

11 x > 99

divide both sides by 11 to isolate x (notice again that 11 is a positive number, so the inequality doesn't change with the division)

x > 9

This inequality is represented by shadowing the right hand side of the number line starting at x = 9. Make sure you draw an EMPTY dot on the number 9 to indicate that x = 9 is NOT included.

4 0
3 years ago
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