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3 years ago

Hi there can anyone help me out with this please that is really good in math please

Mathematics

1 answer:

MariettaO [177]3 years ago

7 0

$826.5 is the present amount at a 10.5% interest rate annually that will lead you to have $1000 in two years.

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Solve for x. <br><br> 3(x + 5) - 2(x + 2) = 20

IRINA_888 [86]

[tex] 3(x + 5) - 2(x + 2) = 20 \\
3x + 15 - 2x - 4 = 20\\
5x + 11 = 20\\
5x = 9\\
x = 1\frac{4}{5} [tex]

4 0

3 years ago

Read 2 more answers

What time is 7 1/2 hours before 2:12 am

Vlad1618 [11]

6:42 pm bc subract 2 hours is 12:12am and then subtract the rest is 7:12pm then you subract 12 minutes is 7 pm the subtract the 18 minutes left is 6:42 pm

4 0

3 years ago

11/10=n/14 Please solve this will give brainliest for the correct answer

Talja [164]

$\text{ Hi there! =)}$

$\large\boxed{n = 15.4}$

$\frac{11}{10} =\frac{n}{14} \\\\\text{ Cross multiply to solve for n:}\\\\11 * 14 = 10 * n\\\\154 = 10n\\\\n = 15.4$

8 0

2 years ago

What is the GCF of <br><br> 2X^3+3X^2+8X+12

tia_tia [17]

Answer:

The GFC is 1

Step-by-step explanation:

5 0

3 years ago

Assume that a cross is made between a heterozygous tall pea plant and a homozygous short pea plant. Fifty offspring are produced

Butoxors [25]

Answer:

1) $\chi^2 = \frac{(30-25)^2}{25} +\frac{(20-25)^2}{25} =1 +1 =2$

2) $df = c-1 = 2-1$

Where c represent the number of categories c=2

Step-by-step explanation:

Previous concepts

A chi-square goodness of fit test "determines if a sample data matches a population".

A chi-square test for independence "compares two variables in a contingency table to see if they are related. In a more general sense, it tests to see whether distributions of categorical variables differ from each another".

Assume the following dataset:

Tall =30 , Short =20

We need to conduct a chi square test in order to check the following hypothesis:

H0: The deviations from a 1:1 ratio (25 tall and 25 short) are due to chance.

H1: The deviations from a 1:1 ratio (25 tall and 25 short) are NOT due to chance.

Part 1

So then we know that the expected values would be 25 for each case

The statistic to check the hypothesis is given by:

$\sum_{i=1}^n \frac{(O_i -E_i)^2}{E_i}$

And if we replace we got:

$\chi^2 = \frac{(30-25)^2}{25} +\frac{(20-25)^2}{25} =1 +1 =2$

Part 2

For this case the degreed of freedom are given by:

$df = c-1 = 2-1$

Where c represent the number of categories c=2

And we can calculate the p value given by:

$p_v = P(\chi^2_{1} >2)=0.157$

And we can find the p value using the following excel code:

"=1-CHISQ.DIST(2,1,TRUE)"

7 0

3 years ago