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3 years ago

15

you were given 15$ from your auntie to go to the mall. you spend 6$ on snacks, and 5$ on shirt on your way to the car,you found

10$ on the ground! what integer shows the amount of money you had when you got home?

1 answer:

Finger [1]3 years ago

5 0

The answer would be 14

Explanation:
If you look on a number line and start from 15 then from there you minus 6 which equals 9 then you minus 5 which equals 4 and add 10 which equals 14 and that 14 is on the positive side of the number line. So 14 is your answer.

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A random sample of 144 recent donations at a certain blood bank reveals that 81 were type A blood. Does this suggest that the ac

Reptile [31]

Answer:

Null hypothesis: $p=0.4$

Alternative hypothesis: $p \neq 0.4$

$z=\frac{0.5625 -0.4}{\sqrt{\frac{0.4(1-0.4)}{144}}}=3.98$

$p_v =2*P(z>3.98)=0.0000689$

Since the p value is very low compared to the significance level we have enough evidence to reject the null hypothesis and we can conclude that the true percent of people with type A of blood is significantly different from 0.4 or 40%

Step-by-step explanation:

Information given

n=144 represent the random sample taken

X=81 represent the number of people with type A blood

$\hat p=\frac{81}{144}=0.5625$ estimated proportion of people with type A blood

$p_o=0.4$ is the value that we want to verify

$\alpha=0.01$ represent the significance level

z would represent the statistic

$p_v{/tex} represent the p value Hypothesis to testWe want to test if the percentage of the population having type A blood is different from 40%.: Null hypothesis:[tex]p=0.4$

Alternative hypothesis: $p \neq 0.4$

the statistic is given by:

$z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}}$ (1)

Replacing the info given we got:

$z=\frac{0.5625 -0.4}{\sqrt{\frac{0.4(1-0.4)}{144}}}=3.98$

Now we can calculate the p value with this probability taking in count the alternative hypothesis:

$p_v =2*P(z>3.98)=0.0000689$

Since the p value is very low compared to the significance level we have enough evidence to reject the null hypothesis and we can conclude that the true percent of people with type A of blood is significantly different from 0.4 or 40%

3 0

2 years ago

Does the expression 1.016x represent a percent increase greater than 12% if the original amount is x? How could I rewrite the ex

GrogVix [38]

Answer:

The expression $1.016x$ does not represent a percent increase greater than 12%.

Step-by-step explanation:

We are asked to find whether the expression $1.016x$ represent a percent increase greater than 12% if the original amount is x.

First of all, we will find 12% increase. The total amount after x% increase would be original amount plus 12% of original amount.

$\text{12 percent increase}=x+\frac{12}{100}x$

$\text{12 percent increase}=x+0.12x$

$\text{12 percent increase}=1.12x$

Since 1.12 is greater than 1.016, therefore, the expression $1.016x$ does not represent a percent increase greater than 12%.

We can rewrite $1.016x$ as:

$1.016x=1x+0.016x=x+0.016x$

Let us convert $0.016$ to percent by multiplying by 100.

$0.016\times 100\%=1.6\%$

Since 1.6% is less than 12%, therefore, the expression $1.016x$ does not represent a percent increase greater than 12%.

4 0

3 years ago

Multiply 3 by b, then multiply c by the result

frez [133]

Hi

Answer:

3bc

Step-by-step explanation:

3×b=3b

3b×c=3bc

5 0

2 years ago

What is the quotient of (x3 - 3x2 + 5x – 3) = (x - 1)?

IRINA_888 [86]

Note: Consider "÷" sign instead of "=".

Given:

$(x^3-3x^2+5x-3)\div (x-1)$

To find:

The quotient.

Solution:

We have,

$(x^3-3x^2+5x-3)\div (x-1)$

It can be written as

$=\dfrac{x^3-3x^2+5x-3}{x-1}$

Splitting the middle terms, we get

$=\dfrac{x^3-x^2-2x^2+2x+3x-3}{x-1}$

$=\dfrac{x^2(x-1)-2x(x-1)+3(x-1)}{x-1}$

Taking out the common factor (x-1), we get

$=\dfrac{(x-1)(x^2-2x+3)}{x-1}$

Cancel the common factors.

$=x^2-2x+3$

The quotient is $x^2-2x+3$ .

Therefore, the correct option is D.

7 0

2 years ago

Read 2 more answers

Wats the equation??<br> 9/4 = Division 4

lara [203]

9/4 = 9 divided by 4

8 0

3 years ago