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3 years ago

10

PLEASE HELP !!!!

1 answer:

Mumz [18]3 years ago

7 0

Answer:

9 hr

Step-by-step explanation:

vary directly means you should think about a constant being multiply to one variable to get another variable.

So we have E=kH

where E=earnings and H=hours and k is the constant of proportionality

So they give us when H=7 we have E=77 giving us 77=k(7) which means the constant of proportionality is 11

So that means no matter what the E and H are k is 11

So we have the equation is E=11H

Replace E with 99 and solve for H

99=11(H)

So H=9

9 hours

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Vicki started jogging the first time she ran she ran 3/16 mile the second time she ran 3/8 mile and the third time she ran 9/16

IrinaK [193]

Answer:

Jogging 6th time.

Step-by-step explanation:

We have been given that Vicki started jogging the first time she ran she ran 3/16 mile the second time she ran 3/8 mile and the third time she ran 9/16 mile.

We can see that the distance Vicki covers each time forms a arithmetic sequence, where 1st term is 3/16.

We know that an arithmetic sequence is in form $a_n=a_1+(n-1)d$ , where,

$a_n$ = nth term of sequence,

$a_1$ = 1st term of sequence,

n = Number of terms in sequence,

d = Common difference.

Let us find common difference of our given sequence as:

$\frac{3}{8}-\frac{3}{16}\Rightarrow \frac{6}{16}-\frac{3}{16}=\frac{3}{16}$

Since Vicki needs to cover more than 1 mile, so we nth term of sequence should be greater than 1.

$1$

Let us solve for n.

$1$

$1$

$1\cdot \frac{16}{3}$

$5.333$

$n>5.333$

We can also write next terms of our sequence as:

$\frac{3}{16},\frac{6}{16}, \frac{9}{16},\frac{12}{16},\frac{15}{16},\frac{18}{16}$

Therefore, Vicki will run more than 1 mile when she is jogging for 6th time.

7 0

3 years ago

Based on historical data, your manager believes that 26% of the company's orders come from first-time customers. A random sample

scoundrel [369]

Answer:

$\hat p \sim N( p, \sqrt{\frac{p (1-p)}{n}})$

And we can use the z score formula given by:

$z = \frac{\hat p -\mu_p}{\sigma_p}$

And if we find the parameters we got:

$\mu_p = 0.26$

$\sigma_p = \sqrt{\frac{0.26(1-0.26)}{158}} = 0.0349$

And we can find the z score for the value of 0.4 and we got:

$z = \frac{0.4-0.26}{0.0349}= 4.0119$

And we can find this probability:

$P(z>4.0119) = 1-P(z$

And if we use the normal standard table or excel we got:

$P(z>4.0119) = 1-P(z$

Step-by-step explanation:

For this case we have the following info given:

$p = 0.26$ represent the proportion of the company's orders come from first-time customers

$n=158$ represent the sample size

And we want to find the following probability:

$p(\hat p >0.4)$

And we can use the normal approximation since we have the following two conditions:

1) np = 158*0.26 = 41.08>10

2) n(1-p) = 158*(1-0.26) = 116.92>10

And for this case the distribution for the sample proportion is given by:

$\hat p \sim N( p, \sqrt{\frac{p (1-p)}{n}})$

And we can use the z score formula given by:

$z = \frac{\hat p -\mu_p}{\sigma_p}$

And if we find the parameters we got:

$\mu_p = 0.26$

$\sigma_p = \sqrt{\frac{0.26(1-0.26)}{158}} = 0.0349$

And we can find the z score for the value of 0.4 and we got:

$z = \frac{0.4-0.26}{0.0349}= 4.0119$

And we can find this probability:

$P(z>4.0119) = 1-P(z$

And if we use the normal standard table or excel we got:

$P(z>4.0119) = 1-P(z$

8 0

2 years ago

Let Q(x, y) be the predicate "If x < y then x2 < y2," with domain for both x and y being R, the set of all real numbers.

Levart [38]

Answer:

a) Q(-2,1) is false

b) Q(-5,2) is false

c)Q(3,8) is true

d)Q(9,10) is true

Step-by-step explanation:

Given data is $Q(x,y)$ is predicate that $x$ then $x^{2}$ . where $x,y$ are rational numbers.

a)

when $x=-2, y=1$

Here $-2$ that is $x$ satisfied. Then

$(-2)^{2}$

$4$ this is wrong. since $4>1$

That is $x^{2}$ $>y^{2}$ Thus $Q(x,y)$ $=Q(-2,1)$ is false.

b)

Assume $Q(x,y)=Q(-5,2)$ .

That is $x=-5, y=2$

Here $-5$ that is $x$ this condition is satisfied.

Then

$(-5)^{2}$

$25$ this is not true. since $25>4$ .

This is similar to the truth value of part (a).

Since in both $x$ satisfied and $x^{2} >y^{2}$ for both the points.

c)

if $Q(x,y)=Q(3,8)$ that is $x=3$ and $y=8$

Here $3$ this satisfies the condition $x$ .

Then $3^{2}$

$9$ This also satisfies the condition $x^{2}$ .

Hence $Q(3,8)$ exists and it is true.

d)

Assume $Q(x,y)=Q(9,10)$

Here $9$ satisfies the condition $x$

Then $9^{2}$

$81$ satisfies the condition $x^{2}$ .

Thus, $Q(9,10)$ point exists and it is true. This satisfies the same values as in part (c)

6 0

3 years ago

Order of Operation

Sphinxa [80]

<em><u>45(x)-20 </u></em>

x=hours

7 hours

x=7

<em><u>45(7)-20 =295</u></em>

4 0

2 years ago

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Jade has a budget of $200

stealth61 [152]

And? It’s there a full question?

8 0

3 years ago