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3 years ago

Which numbers below are solutions to the inequality 4y+3≤−7?

Mathematics

1 answer:

MariettaO [177]3 years ago

4 0

Answer:

answer is in picture

Step-by-step explanation:

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If you rearrange the equation so w is the independent variable, then what is u? -2u+6w=9

Serjik [45]

Answer:

u = - ( 9-6w)/2

Step-by-step explanation:

to evaluate for u in the expression -2u+6w=9 is simply to look for a way such that we would express u in terms of w and other variables.

solution

-2u+6w=9

-2u = 9 - 6w

divide both sides by the coefficient of u which is -2

-2u/u = 9 - 6w/-2

u = - ( 9-6w)/2

therefore the value of u when rearranged in the equation -2u+6w=9 is evaluated to be equals to

u = - ( 9-6w)/2

7 0

3 years ago

A scuba diver used the expression below to describe his position in relation to sea level.

alex41 [277]

It is the first statement. It is positive two, so it’s above sea level. We can immediately get rid of the last two statements. Then, it is negative 20. The is 18 feet below sea level, because 2-20 is -18. Positive 8 shows rising up by 8 feet. The correct statement is the first one.

3 0

3 years ago

Read 2 more answers

After breaking your penny bank, you notice that you have $4.87 in quarters ($0.25), dimes ($0.10), nickels ($0.05), and pennies

ira [324]

Answer: Option 'c' is correct.

Step-by-step explanation:

Let the number of dimes be 'x'.

Let the number of nickels be ' $\dfrac{3}{2}x$ '.

Let the number of pennies be $\dfrac{3}{2}\times 2x=3x$

Let the number of quarters be '3x-34'.

As we know that

1 quarter = $0.25

1 dime = $0.10

1 nickel = $0.05

1 penny = $0.01

According to question, we get that

$0.10x+0.05\times \dfrac{3x}{2}+0.01\times 3x+0.25(3x-34)=4.87\\\\0.10x+0.075x+0.03x+0.75x-8.5=4.87\\\\\\0.955x=4.87+8.5\\\\0.955x=13.37\\\\x=\dfrac{13.37}{0.955}\\\\x=14$

So, the number of pennies is given by

$3x=3\times 14=42$

Hence, Option 'c' is correct.

3 0

3 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

the cost of belonging to a gym can be modeled by by c(m)=50m+79.50 what is the meaning of the slope and y-intercept

Bess [88]

The slope is the relationship between the cost of the gym and how long you are there for.
the y-intercept is how much it costs to go to the gym for no hours ie the start-up cost.

5 0

3 years ago