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Blizzard [7]
3 years ago
11

Chris received a $90.00 check for his birthday. The check was for 5 more than twice the amount of the check that he got last yea

r for his birthday. How much money did Chris receive last year for his birthday? (1 point) $85.00 $42.50 $50.50 $47.50
Mathematics
2 answers:
kobusy [5.1K]3 years ago
8 0
The easiest way to figure out a simpler math problem like this is to look at all of the answers.

Look at all of the answers and see which one times 2 is 5 less than 90.

A.) 85.00 ---->  85 x 2 = 170 (Nope)
B.) 42.50 ----> 42.50 x 2 = 85 (Yep)
C.) 50.50 ----> 50.50 x 2 = 101 (Nope)
D.) 47.50 ----> 47.50 x 2 = 95 (Nope)

That means the correct answer is $42.50

Hope this helps!
vodomira [7]3 years ago
3 0
The answer to this question is 42.50
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During the first hour of operation the farm stand owner sold 15% of the apples he had. On the second hour he sold 20% of the rem
zzz [600]

Answer:

38.25 kg

Step-by-step explanation:

Let the number of apples sold = x

Now, on the first hour he sold 15% of the apples i.e. he sold 0.15x kg of apples.

So, the remaining apples = x - 0.15x = 0.85x

Further, on the second hour he sold 20% of the remaining apples i.e. he sold 0.20 × 0.85x = 0.17x kg of apples.

Thus, the remaining apples = 0.85x - 0.17x = 0.68x

Then, on the third hour he sold 25% of the remaining apples i.e. he sold 0.25 × 0.68x = 0.17x kg of apples.

So, the remaining apples = 0.68x - 0.17x = 0.51x

Moreover, on the fourth hour he will sell 30% ( i.e. 5% more of 25% ) of the remaining apples i.e. he sold 0.30 × 0.51x = 0.153x kg of apples.

As, during the second hour, he sold 25.5 kg of apples.

Then, 0.17x = 25.5 i.e. x=\frac{25.5}{0.17} i.e. x = 250

Hence, during the fourth hour, he will sell 0.153x = 0.153 × 250 = 38.25 kg of apples.

7 0
3 years ago
Which inequality is false? A. 4 < 9 B. –6 < –3 C. –5 > –2 D. 12 > –7
CaHeK987 [17]
Simple...

which inequality is false? 

A.) 4<9 (True)

B.) -6<-3 (True)

C.) -5>-2 (False)

D.) 12>-7 (True)

Thus, your answer.
4 0
3 years ago
Read 2 more answers
Write the equation of a line that passes through two points (-1, 0) and (4, -5).
sleet_krkn [62]

Answer:

<u>y = -1 (x) -1</u>

Step-by-step explanation:

If you plug in the numbers from the coordinates into the equation, you will find that it's true.

-1 is x and 0 is y. Plug in the #s like so:

y = -1 (x) -1

0 = -1 (-1) - 1

0 = 1 - 1

0 = 0

Therefore, the equation above represents the line that passes through both these points.

8 0
3 years ago
What is the factor tree of 8
8090 [49]
THE FACTOR TREE IS : 8=2^3
8 0
3 years ago
In a simple random sample of 14001400 young​ people, 9090​% had earned a high school diploma. Complete parts a through d below.
ratelena [41]

Answer:

(a) The standard error is 0.0080.

(b) The margin of error is 1.6%.

(c) The 95% confidence interval for the percentage of all young people who earned a high school diploma is (88.4%, 91.6%).

(d) The percentage of young people who earn high school diplomas has ​increased.

Step-by-step explanation:

Let <em>p</em> = proportion of young people who had earned a high school diploma.

A sample of <em>n</em> = 1400 young people are selected.

The sample proportion of young people who had earned a high school diploma is:

\hat p=0.90

(a)

The standard error for the estimate of the percentage of all young people who earned a high school​ diploma is given by:

SE_{\hat p}=\sqrt{\frac{\hat p(1-\hat p)}{n}}

Compute the standard error value as follows:

SE_{\hat p}=\sqrt{\frac{\hat p(1-\hat p)}{n}}

       =\sqrt{\frac{0.90(1-0.90)}{1400}}\\

       =0.008

Thus, the standard error for the estimate of the percentage of all young people who earned a high school​ diploma is 0.0080.

(b)

The margin of error for (1 - <em>α</em>)% confidence interval for population proportion is:

MOE=z_{\alpha/2}\times SE_{\hat p}

Compute the critical value of <em>z</em> for 95% confidence level as follows:

z_{\alpha/2}=z_{0.05/2}=z_{0.025}=1.96

Compute the margin of error as follows:

MOE=z_{\alpha/2}\times SE_{\hat p}

          =1.96\times 0.0080\\=0.01568\\\approx1.6\%

Thus, the margin of error is 1.6%.

(c)

Compute the 95% confidence interval for population proportion as follows:

CI=\hat p\pm MOE\\=0.90\pm 0.016\\=(0.884, 0.916)\\\approx (88.4\%,\ 91.6\%)

Thus, the 95% confidence interval for the percentage of all young people who earned a high school diploma is (88.4%, 91.6%).

(d)

To test whether the percentage of young people who earn high school diplomas has​ increased, the hypothesis is defined as:

<em>H₀</em>: The percentage of young people who earn high school diplomas has not​ increased, i.e. <em>p</em> = 0.80.

<em>Hₐ</em>: The percentage of young people who earn high school diplomas has not​ increased, i.e. <em>p</em> > 0.80.

Decision rule:

If the 95% confidence interval for proportions consists the null value, i.e. 0.80, then the null hypothesis will not be rejected and vice-versa.

The 95% confidence interval for the percentage of all young people who earned a high school diploma is (88.4%, 91.6%).

The confidence interval does not consist the null value of <em>p</em>, i.e. 0.80.

Thus, the null hypothesis is rejected.

Hence, it can be concluded that the percentage of young people who earn high school diplomas has ​increased.

8 0
3 years ago
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