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3 years ago

In 17, 874,000 what is the value of 1

Mathematics

1 answer:

artcher [175]3 years ago

4 0

Answer:

<h2>Ten Millions</h2>

Step-by-step explanation:

1 = Ten Millions

7 = Millions

8 = Hundred Thousands

7 = Ten Thousands

4 = Thousands

0 = Hundreds

0 = Tens

0 = Ones

Hope this helps! <3

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Find the P-value for the indicated hypothesis test with the given standardized test statistic, z. Decide whether to reject H0

Leviafan [203]

Answer: We do not reject the null hypothesis.

Step-by-step explanation:

When the p-value is greater than the significance level , then we do not reject the null hypothesis or if p-value is smaller than the significance level , then we reject the null hypothesis.

Given : Test statistic : $z = -2.28$

Significance level : $\alpha=0.02$

By using the standard normal distribution table ,

The p-value corresponds to the given test statistic ( two tailed ):-

$2P(Z>|z|)=2P(Z>2.28)=0.0226$

Since the p-value is greater than the significance level of 0.02.

Then , we do not reject the null hypothesis.

8 0

3 years ago

Sketch the graph of each line y=-9x-5

ycow [4]

I can’t answer it without a picture of the graph

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2 years ago

Miss Sandy is fencing in her rectangular backyard which is 3x^2 feet wide and 4xy + 5x^3 y^2 feet long. Find perimeter of yard H

hjlf

Given:

Width of rectangular backyard = $3x^2$ feet

Length of rectangular backyard = $4xy+5x^3y^2$ feet

To find:

The perimeter of yard.

Solution:

Formula or perimeter of a rectangle is

$Perimeter=2(length +width)$

Using this formula, the perimeter of the rectangular backyard is

$Perimeter=2(4xy+5x^3y^2+3x^2)$

$Perimeter=8xy+10x^3y^2+6x^2$

Therefore, the perimeter of the rectangular backyard is $8xy+10x^3y^2+6x^2$ feet. So, $8xy+10x^3y^2+6x^2$ feet fencing is needed.

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2 years ago

Help!!! Photo is linked below.

Gala2k [10]

28.8 degrees

tan-1 = 71/81

4 0

1 year ago

TASK 3= 50 POINTSSS

Nadusha1986 [10]

$After \: 1 \: year:25400 + 25400 \times 11\% = 25400 + 25400 \times 0.11 = 25400(1 + 0.11) = 25400 \times 1.11 \\ After \: 2 \: year:25400 \times {1.11}^{2} \\ After \: x \: year:25400 \times {1.11}^{x} \\ We\: have\: an\: inequality:25400 \times {1.11}^{x} > 501170 \\ \Leftrightarrow {1.11}^{x} > \frac{50117}{2540}\\ \Leftrightarrow x > \log _{1.11}(\frac{50117}{2540}) \\ \Leftrightarrow x>29\\\Rightarrow After\: 29\: years$

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2 years ago