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nadya68 [22]
3 years ago
11

Hello if you're able to answer this question help me and provide work as well, Thank you.

Mathematics
1 answer:
White raven [17]3 years ago
6 0

Answer:

answer is no. a -1/8 for sure

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Which equations have the same value of x as 3/5 (30x-15) = 72? Select three options.
AfilCa [17]
I got:
c) 18x-9=72
d) 3(6x-3)=72
e) x=4.5
If you need any explanations for the answers, please tell me!
7 0
3 years ago
Read 2 more answers
Question 26 pleaseee
LuckyWell [14K]

Answer:

The inequality is 55+10x\leq 105

The greatest length of time Jeremy can rent the jet ski is 5 and Jeremy can rent maximum of 135 minutes.

Step-by-step explanation:

Given: Cost of first hour rent of jet ski is $55

          Cost of each additional 15 minutes of jet ski is $10

          Jeremy can spend no more than $105

Assuming the number of additional 15-minutes increment be "x"

Jeremy´s total spending would be first hour rental fees and additional charges for each 15-minutes of jet ski.

Lets put up an expression for total spending of Jeremy.

\$55+ \$ 10\times x

We also know that Jeremy can not spend more than $105

∴ Putting up the total spending of Jeremy in an inequality.

\$ 55+\$10x\leq \$ 105

Now solving the inequality to find the greatest number of time Jeremy can rent the jet ski,

⇒ \$ 55+\$10x\leq \$105

Subtracting both side by 55

⇒ \$ 10x\leq \$50

Dividing both side by 10

⇒x\leq \frac{50}{10}

∴ x\leq 5

Therefore, Jeremy can rent for 60\ minutes + 5\times 15\\= 60\ minutes + 75= 135\ minutes

Jeremy can rent maximum of 135 minutes.

5 0
3 years ago
the pictograph below shows number of packages of different brand of batteries at a store sold on Fridaywhich statement is best s
Wewaii [24]
I am happy to help... but <span>Please be more specific with your question. </span>
8 0
3 years ago
The length of a photograph is 1 cm less than twice the width. The area is 91cm^2. find the width and length.
maksim [4K]

Answer:

Width=6.5 cm

Length=12 cm

Step-by-step explanation:

Step 1: Express the lengths and widths

Width=w

Length=l, but 1 cm less than twice the width=(2×w)-1=2 w-1

Step 2: Solve for the length and width

A=L×W

where;

A=area of the photograph

L=length of the photograph

W=width of the photograph

In our case;

A=91 cm²

L=2 w-1

W=w

91=(2 w-1)w

2 w²-w=91

2 w²-w-91=0, is a quadratic equation

solve for w

w={-1±√(-1²-4×2×-91)}/(2×2)

w=(-1±27)/4

w=(27-1)/4=6.5, or (-1-27)/4=-8

Take w=6.5 cm

L=(2×6.5)-1=13-1=12 cm

Width=6.5 cm

Length=12 cm

6 0
3 years ago
When we toss a coin, there are two possible outcomes: a head or a tail. Suppose that we toss a coin 100 times. Estimate the appr
marin [14]

Answer:

96.42% probability that the number of tails is between 40 and 60.

Step-by-step explanation:

I am going to use the binomial approximation to the normal to solve this question.

Binomial probability distribution

Probability of exactly x sucesses on n repeated trials, with p probability.

Can be approximated to a normal distribution, using the expected value and the standard deviation.

The expected value of the binomial distribution is:

E(X) = np

The standard deviation of the binomial distribution is:

\sqrt{V(X)} = \sqrt{np(1-p)}

Normal probability distribution

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the zscore of a measure X is given by:

Z = \frac{X - \mu}{\sigma}

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

When we are approximating a binomial distribution to a normal one, we have that \mu = E(X), \sigma = \sqrt{V(X)}.

In this problem, we have that:

100 tosses, so n = 100

Two outcomes, both equally as likely. So p = \frac{1}{2} = 0.5

So

E(X) = np = 100*0.5 = 50

\sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{100*0.5*0.5} = 5

Estimate the approximate probability that the number of tails is between 40 and 60.

Using continuity correction.

P(40 - 0.5 \leq X \leq 60 + 0.5) = P(39.5 \leq X \leq 60.5)

This is the pvalue of Z when X = 60.5 subtracted by the pvalue of Z when X = 39.5. So

X = 60.5

Z = \frac{X - \mu}{\sigma}

Z = \frac{60.5 - 50}{5}

Z = 2.1

Z = 2.1 has a pvalue of 0.9821

X = 39.5

Z = \frac{X - \mu}{\sigma}

Z = \frac{39.5 - 50}{5}

Z = -2.1

Z = -2.1 has a pvalue of 0.0179

0.9821 - 0.0179 = 0.9642

96.42% probability that the number of tails is between 40 and 60.

8 0
3 years ago
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