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1 year ago

4.what is the number in standard form? 7.053×10^6

Mathematics

2 answers:

vichka [17]1 year ago

8 0

The answer is 7,053,000.

Vikki [24]1 year ago

5 0

Answer:

The standard form is $7.053\times 10^6=7053000$

Step-by-step explanation:

Given : Expression $7.053\times 10^6$

To find : What is the number in standard form ?

Solution :

Standard form is the way of writing a number in expanded or simpler form.

Number $7.053\times 10^6$

Remove decimal and open power of 10.

$7.053\times 10^6=\frac{7053}{1000}\times 1000000$

$7.053\times 10^6=7053\times 1000$

$7.053\times 10^6=7053000$

Therefore, The standard form is $7.053\times 10^6=7053000$

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Data show that men between the ages of 20 and 29 in a general population have a mean height of 69.3 inches, with a standard dev

dangina [55]

Data show that men between the ages of 20 and 29 in a general population have a mean height of 69.3 inches, with a standard deviation of 2.5 inches. a baseball analyst wonders whether the standard deviation of heights of major-league baseball players is less than 2.5 inches. the heights (in inches) of 20 randomly selected players are shown in the table.

72 74 71 72 76

70 77 75 72 72

77 72 75 70 73

74 75 73 74 74

What are the correct hypotheses for this test?

The null hypothesis is H₀?: ____ 2.5

The alternative hypothesis is H₁?: ____ 2.5

Calculate the value of the test statistic.

x² = _____ (Round to three decimal places)

Answer:

Null hypothesis, H₀: σ = 2.5

Alternative hypothesis, Hₐ: μ<2.5

Test statistic = 12.920

Step-by-step explanation:

Given Data shows that:

men between the ages of 20 and 29 in a general population have a mean height of 69.3 inches, with a standard deviation of 2.5 inches

We consider a random sample of 20 selected baseball players.

Therefore;

The Null and Alternative hypothesis are as follows:

The Null hypothesis is the standard deviation of the heights of major league baseball players is not less than 2.5 inches.

Null hypothesis, H₀: σ = 2.5

On the other hand: The Alternative hypothesis is the standard deviation of the heights of major league baseball players is less than 2.5 inches.

Alternative hypothesis, Hₐ: μ<2.5

The Mean Calculation is:

$\bar{x} = \frac{1}{2} \sum x_i$

$= \frac{1}{20} (72+74+...+74) \\ \\ = \frac{1468}{20} \\ \\ =73.4$

The sample standard deviation is:

$s = \sqrt{\frac{1}{n-1} \sum (x_1 - \bar{x})^2 }$

$= \sqrt{\frac{1}{20-1} \sum (72-73.4)^2 + ...+(74-73.4)^2 } \\ \\ = \sqrt{4.25} \\ \\ = 2.06$

The test statistics is now determined as :

$x^2 = \frac{(n-1)s^2}{\sigma^2} \\ \\ = \frac{(20-1)(2.06)^2}{(2.5)^2} \\ \\ = \frac{19*4.25}{6.25} \\ \\ = \frac{80.75}{6.25} \\ \\ = 12.920$

4 0

2 years ago

The areas of the squares created by the side lengths of the triangle are shown. Which best explains whether this triangle is a r

nikitadnepr [17]

Answer:

Based on the converse of the Pythagorean Theorem, the triangle is not a right triangle, because $9+12\neq 15$

Step-by-step explanation:

The complete question in the attached figure

we know that

If the length sides of a triangle, satisfy the Pythagorean Theorem, then is a right triangle

$c^2=a^2+b^2$

where

c is the hypotenuse (the greater side)

a and b are the legs

In this problem

The length sides squared of the triangle are equal to the areas of the squares

$c^2=15\ in^2$

$a^2=12\ in^2$

$b^2=9\ in^2$

substitute

$15=12+9$

$15=21$ ----> is not true

The length sides not satisfy the Pythagorean Theorem

therefore

Based on the converse of the Pythagorean Theorem, the triangle is not a right triangle, because $9+12\neq 15$

8 0

2 years ago

Read 2 more answers

A statistics instructor who teaches a lecture section of 160 students wants to determine whether students have more difficulty w

Leona [35]

Answer:

There is no evidence that there is no significant difference between the sample means

Step-by-step explanation:

given that a statistics instructor who teaches a lecture section of 160 students wants to determine whether students have more difficulty with one-tailed hypothesis tests or with two-tailed hypothesis tests. On the next exam, 80 of the students, chosen at random, get a version of the exam with a 10-point question that requires a one-tailed test. The other 80 students get a question that is identical except that it requires a two-tailed test. The one-tailed students average 7.81 points, and their standard deviation is 1.06 points

The two-tailed students average 7.64 points, and their standard deviation is 1.33 points.

Group One tailed X Two tailed Y

Mean 7.8100 7.6400

SD 1.0600 1.3300

SEM 0.1185 0.1487

N 80 80

$H_0:\bar x=\bar y\\H_a: \bar x \neq \bar y$

(Two tailed test)

The mean of One tailed X minus Two tailed Y equals 0.1700

t = 0.8940

df = 158

p value =0.3727

p is greater than alpha 0.05

There is no evidence that there is no significant difference between the sample means

3 0

2 years ago

A building has an entry the shape of a parabolic arch 74 ft high and 28 ft wide at the base as shown below.

sergiy2304 [10]

The equation of a parabola with vertex at (h,k) is
y=a(x-h)²+k

vertex isi at (0,0)

y=a(x-0)²+0
y=a(x)²
y=ax²
find a

we see that one point is (14,-74)
x=14 and y=-74

-74=a(14²)
-74=196a
divide both sides by 196
-37/98=a

the equation is

$y= \frac{-37}{98} x^2$

6 0

2 years ago

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In isosceles △ABC (AC = BC) with base angle 30° CD is a median. How long is the leg of △ABC, if sum of the perimeters of △ACD an

SIZIF [17.4K]

Note necessary facts about isosceles triangle ABC:

The median CD drawn to the base AB is also an altitude to tha base in isosceles triangle (CD⊥AB). This gives you that triangles ACD and BCD are congruent right triangles with hypotenuses AC and BC, respectively.
The legs AB and BC of isosceles triangle ABC are congruent, AC=BC.
Angles at the base AB are congruent, m∠A=m∠B=30°.

1. Consider right triangle ACD. The adjacent angle to the leg AD is 30°, so the hypotenuse AC is twice the opposite leg CD to the angle A.

AC=2CD.

2. Consider right triangle BCD. The adjacent angle to the leg BD is 30°, so the hypotenuse BC is twice the opposite leg CD to the angle B.

BC=2CD.

3. Find the perimeters of triangles ACD, BCD and ABC:

$P_{ACD}=AC+CD+AD=2CD+CD+AD=3CD+AD;$

$P_{BCD}=BC+CD+BD=2CD+CD+AD=3CD+AD;$

$P_{ABC}=AC+BC+AB=2CD+2CD+AD+BD=4CD+2AD.$

4. If sum of the perimeters of △ACD and △BCD is 20 cm more than the perimeter of △ABC, then

$P_{ACD}+P_{BCD}=P_{ABC}+20,\\ \\3CD+AD+3CD+AD=4CD+2AD+20,\\ \\6CD+2AD=4CD+2AD+20,\\ \\2CD=20.$

5. Since AC=BC=2CD, then the legs AC and BC of isosceles triangles have length 20 cm.

Answer: 20 cm.

8 0

2 years ago

Read 2 more answers