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3 years ago

What is 6/5 x 10 please help

Mathematics

2 answers:

Jlenok [28]3 years ago

8 0

Answer:

Step-by-step explanation:

6 x 10= 60 60/5 = 12

inessss [21]3 years ago

5 0

Answer:

Step-by-step explanation:

easy peasy

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I don't even know where to start.

alexira [117]

Don't know if this helps but the small thing will fit in the big shape about 6 times

7 0

3 years ago

Simplify the following 4√28÷3√7

finlep [7]

Answer:

8/3 ( Fraction Form), 2.6 ( Decimal Form), 2 2/3 ( Mixed Number Form)

Step-by-step explanation:

Simplify the radical by breaking the radicand up into a product of known factors, assuming positive real numbers.

6 0

3 years ago

Read 2 more answers

⦁ In a simple random sample of 1219 US adults, 354 said that their favorite sport to watch is football. Construct a 95% confiden

fomenos

Answer:

95% confidence interval for the proportion of adults in the United States whose favorite sport to watch is football is [0.265 , 0.316].

Step-by-step explanation:

We are given that in a simple random sample of 1219 US adults, 354 said that their favorite sport to watch is football.

Firstly, the pivotal quantity for 95% confidence interval for the proportion of adults in the United States whose favorite sport to watch is football is given by;

P.Q. = $\frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ~ N(0,1)

where, $\hat p$ = proportion of adults in the United States whose favorite sport to watch is football in a sample of 1219 adults = $\frac{354}{1219}$

n = sample of US adults = 1291

p = population proportion of adults

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

So, 95% confidence interval for the population proportion, p is ;

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

significance level are -1.96 & 1.96}

P(-1.96 < $\frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ < 1.96) = 0.95

P( $-1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} }$ < ${\hat p-p}$ < $1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} }$ ) = 0.95

P( $\hat p-1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} }$ < p < $\hat p+1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} }$ ) = 0.95

95% confidence interval for p = [ $\hat p-1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} }$ , $\hat p+1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} }$ ]

= [ $\frac{354}{1219}-1.96 \times {\sqrt{\frac{\frac{354}{1219}(1-\frac{354}{1219})}{1219} }$ , $\frac{354}{1219}+1.96 \times {\sqrt{\frac{\frac{354}{1219}(1-\frac{354}{1219})}{1219} }$ ]

= [0.265 , 0.316]

Therefore, 95% confidence interval for the proportion of adults in the United States whose favorite sport to watch is football is [0.265 , 0.316].

5 0

3 years ago

How many extraneous solutions does the equation below have? StartFraction 9 Over n squared 1 EndFraction = StartFraction n 3 Ove

BigorU [14]

The equation has one extraneous solution which is n ≈ 2.38450287.

Given that,

The equation;

$\dfrac{9}{n^2+1} =\dfrac{n+3}{4}$

We have to find,

How many extraneous solutions does the equation?

According to the question,

An extraneous solution is a solution value of the variable in the equations, that is found by solving the given equation algebraically but it is not a solution of the given equation.

To solve the equation cross multiplication process is applied following all the steps given below.

$\rm \dfrac{9}{n^2+1} =\dfrac{n+3}{4}\\\\9 (4) = (n+3) (n^2+1)\\\\36 = n(n^2+1) + 3 (n^2+1)\\\\36 = n^3+ n + 3n^2+3\\\\n^3+ n + 3n^2+3 - 36=0\\\\n^3+ 3n^2+n -33=0\\$