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Natali5045456 [20]
2 years ago
11

Mary won £5000 in a competition

Mathematics
1 answer:
Ahat [919]2 years ago
5 0

Answer: £833

Step-by-step explanation:

1. Multiply 279 by 9 (equals 2,511)

2. Multiply 184 by 9 (1,656)

3. Add 2,511 and 1,656 together (4,167)

4. Subtract 4,167 from 5,000 (833)

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What is y+2=-4(X-6) in Ax+By=C form
PSYCHO15rus [73]
Y+2=-4x+24
2=-4x-y+24
-22=-4x-y
7 0
3 years ago
The decimal d is formed by writing in succession all the positive integers in increasing order after the decimal point; that is,
Maksim231197 [3]

Answer:

2

Step-by-step explanation:

We are given that a decimal number  d

d=0.12345678...

Where d is formed by writing in succession all the positive integers in increasing order after decimal point.

We have to find the 100th digit of d to the right of the decimal point.

Place of first digit 1 after decimal=Tenth

Place of second digit 2 after decimal=Hundredth

Place of third digit  3 after decimal=Thousandth

Therefore, 100th digit of d to the right of the decimal point=2

5 0
3 years ago
() + -8= -6 what is the awnser
Mariana [72]

Answer:x=2

Step-by-step explanation:

change () to x

x + -8= -6

-6+8=2

x=2

5 0
3 years ago
One parent is a cystic fibrosis carrier, and the other has
Marysya12 [62]

Answer:

A - 0%

B- 50%

C- 50%

D- 100%

Step-by-step explanation:

Cystic fibrosis is inherited in an autosomal recessive form, meaning that a person has to inherit two abnormal genes for the disease to manifest. In the case of this question, one parent is a gene carrier, so his genotype is Aa, while the other does not have the cystic fibrosis gene, so AA.

Performing the cross of Aa x AA, we can see that:

a.) The probability of a child would have cystic fibrosis is 0%, since the disease is recessive and to be affected it should receive a recessive gene from each parent.

b.) The probability of a child would be a carrier is 50%, as 50% of the crossing phenotypes are Aa.

c.) The probability of a child would not have cystic fibrosis and is not a carrier is 50%, as 50% of the child's genotype is AA.

d.) The probability of a child would be healthy is 100%, as of all possible phenotypes none is affected.

5 0
3 years ago
The amount of lateral expansion (mils) was determined for a sample of n = 8 pulsed-power gas metal arc welds used in LNG ship co
Alona [7]

Answer:

95% Confidence interval for the variance:

3.6511\leq \sigma^2\leq 34.5972

95% Confidence interval for the standard deviation:

1.9108\leq \sigma \leq 5.8819

Step-by-step explanation:

We have to calculate a 95% confidence interval for the standard deviation σ and the variance σ².

The sample, of size n=8, has a standard deviation of s=2.89 miles.

Then, the variance of the sample is

s^2=2.89^2=8.3521

The confidence interval for the variance is:

\dfrac{ (n - 1) s^2}{ \chi_{\alpha/2}^2} \leq \sigma^2 \leq \dfrac{ (n - 1) s^2}{\chi_{1-\alpha/2}^2}

The critical values for the Chi-square distribution for a 95% confidence (α=0.05) interval are:

\chi_{0.025}=1.6899\\\\\chi_{0.975}=16.0128

Then, the confidence interval can be calculated as:

\dfrac{ (8 - 1) 8.3521}{ 16.0128} \leq \sigma^2 \leq \dfrac{ (8 - 1) 8.3521}{1.6899}\\\\\\3.6511\leq \sigma^2\leq 34.5972

If we calculate the square root for each bound we will have the confidence interval for the standard deviation:

\sqrt{3.6511}\leq \sigma\leq \sqrt{34.5972}\\\\\\1.9108\leq \sigma \leq 5.8819

6 0
3 years ago
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