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3 years ago

Will the number of 1/4 × 6 be less than 6 or greater than 6

2 answers:

8 0

Hello! the answer would be,
Less then 6 because 1/4*6 is 6/4 which is 1 and 2/4 then 1 and 1/2
Hope I've helped and ask questions if you're confused!
Sincerely, Kaylie :)

LuckyWell [14K]3 years ago

7 0

The answer is less then six

You might be interested in

Which statement is true?

defon

Answer:

b. Multiplying decimals less than one is finding a part of a part.

Step-by-step explanation:

A part of a given number implies a selected part of a given number which can be expressed as a fraction of the number. Example, the half of any number is $\frac{1}{2}$ .

The product of a part and another part can be expressed as;

$\frac{1}{3}$ x $\frac{2}{3}$ = $\frac{2}{9}$

or,

$\frac{1}{2}$ x $\frac{1}{2}$ = $\frac{1}{4}$

or,

$\frac{1}{2}$ x $\frac{2}{3}$ = $\frac{2}{6}$ = $\frac{1}{3}$

The three examples shows: a part x a part.

Therefore, it would be observed that multiplying decimals less than one is the same as finding a part of a part.

5 0

3 years ago

Question

Fantom [35]

Answer:

a. More than one pie

Step-by-step explanation:

2/3 + 3/4

8/12 + 9/12 = 17/12

17/12 = 1 5/12

3 0

3 years ago

What are the possible values of x if (4x – 5)2 = 49?

STatiana [176]

2(4x - 5) = 49
8x - 10 = 49
8x - 10 + 10 = 49 + 10
8x = 59
8x/8 = 59/8
x = 7.375

3 0

2 years ago

999999999998÷30×536+9999999

Anettt [7]

Answer:

17866676666629.933594

Step-by-step explanation:

because oBaMa

4 0

2 years ago

A certain geneticist is interested in the proportion of males and females in the population who have a minor blood disorder. In

lord [1]

Answer:

95% confidence interval for the difference between the proportions of males and females who have the blood disorder is [-0.064 , 0.014].

Step-by-step explanation:

We are given that a certain geneticist is interested in the proportion of males and females in the population who have a minor blood disorder.

A random sample of 1000 males, 250 are found to be afflicted, whereas 275 of 1000 females tested appear to have the disorder.

Firstly, the pivotal quantity for 95% confidence interval for the difference between population proportion is given by;

P.Q. = $\frac{(\hat p_1-\hat p_2)-(p_1-p_2)}{\sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1}+ \frac{\hat p_2(1-\hat p_2)}{n_2}} }$ ~ N(0,1)

where, $\hat p_1$ = sample proportion of males having blood disorder= $\frac{250}{1000}$ = 0.25

$\hat p_2$ = sample proportion of females having blood disorder = $\frac{275}{1000}$ = 0.275

$n_1$ = sample of males = 1000

$n_2$ = sample of females = 1000

$p_1$ = population proportion of males having blood disorder

$p_2$ = population proportion of females having blood disorder

Here for constructing 95% confidence interval we have used Two-sample z proportion statistics.

So, 95% confidence interval for the difference between the population proportions, ( $p_1-p_2$ ) is ;

P(-1.96 < N(0,1) < 1.96) = 0.95 {As the critical value of z at 2.5% level

of significance are -1.96 & 1.96}

P(-1.96 < $\frac{(\hat p_1-\hat p_2)-(p_1-p_2)}{\sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1}+ \frac{\hat p_2(1-\hat p_2)}{n_2}} }$ < 1.96) = 0.95

P( $-1.96 \times {\sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1}+ \frac{\hat p_2(1-\hat p_2)}{n_2}} }$ < ${(\hat p_1-\hat p_2)-(p_1-p_2)}$ < $1.96 \times {\sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1}+ \frac{\hat p_2(1-\hat p_2)}{n_2}} }$ ) = 0.95

P( $(\hat p_1-\hat p_2)-1.96 \times {\sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1}+ \frac{\hat p_2(1-\hat p_2)}{n_2}} }$ < ( $p_1-p_2$ ) < $(\hat p_1-\hat p_2)+1.96 \times {\sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1}+ \frac{\hat p_2(1-\hat p_2)}{n_2}} }$ ) = 0.95

95% confidence interval for ( $p_1-p_2$ ) =

[ $(\hat p_1-\hat p_2)-1.96 \times {\sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1}+ \frac{\hat p_2(1-\hat p_2)}{n_2}} }$ , $(\hat p_1-\hat p_2)+1.96 \times {\sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1}+ \frac{\hat p_2(1-\hat p_2)}{n_2}} }$ ]

= [ $(0.25-0.275)-1.96 \times {\sqrt{\frac{0.25(1-0.25)}{1000}+ \frac{0.275(1-0.275)}{1000}} }$ , $(0.25-0.275)+1.96 \times {\sqrt{\frac{0.25(1-0.25)}{1000}+ \frac{0.275(1-0.275)}{1000}} }$ ]

= [-0.064 , 0.014]

Therefore, 95% confidence interval for the difference between the proportions of males and females who have the blood disorder is [-0.064 , 0.014].

8 0

3 years ago