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2 years ago

14

Mario has a 500 page novel which he is required to read over summer break for his upcoming language arts class if Mario read 24

page each day write a function to represent the number of pages that Mario left to read p after d days

1 answer:

KonstantinChe [14]2 years ago

3 0

Answer:

p(d) = 24p + 500d

Step-by-step explanation:

p(d) = 24p + 500d

hopefully this helps ;)

Can you answer my question pleaseeee?

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Arthur wrote that 15 – 14.7 = 3.

PilotLPTM [1.2K]

3 is incorrect because 14.7 + 3 = 17.7
The answer of 15 - 14.7 = 0.3

8 0

1 year ago

The box which measures 70cm X 36cm X 12cm is to be covered by a canvas. How many meters of canvas of width 80cm would be require

grigory [225]

Answer:

142.2 meters.

Step-by-step explanation:

We have been given that a box measures 70 cm X 36 cm X 12 cm is to be covered by a canvas.

Let us find total surface area of box using surface area formula of cuboid.

$\text{Total surface area of cuboid}=2(lb+bh+hl)$ , where,

$l$ = Length of cuboid,

$b$ = Breadth of cuboid,

$w$ = Width of cuboid.

$\text{Total surface area of box}=2(70\cdot36+36\cdot 12+12\cdot 70)$

$\text{Total surface area of box}=2(2520+432+840)$

$\text{Total surface area of box}=2(3792)$

$\text{Total surface area of box}=7584$

Therefore, the total surface area of box will be 7584 square cm.

To find the length of canvas that will cover 150 boxes, we will divide total surface area of 150 such boxes by width of canvass as total surface area of canvas will also be the same.

$\text{Width of canvas* Length of canvass}=\text{Total surface area of 150 boxes}$

$80\text{ cm}\times\text{ Length of canvass}=150\times 7584\text{cm}^2$

$\text{ Length of canvass}=\frac{150\times 7584\text{ cm}^2}{80\text{ cm}}$

$\text{ Length of canvass}=\frac{1137600\text{ cm}^2}{80\text{ cm}}$

$\text{ Length of canvass}=14220\text{ cm}$

Let us convert the length of canvas into meters by dividing 14220 by 100 as 1 meter equals to 100 cm.

$\text{ Length of canvass}=\frac{14220\text{ cm}}{100\frac{cm}{m}}$

$\text{ Length of canvass}=\frac{14220\text{ cm}}{100\frac{cm}{m}}$

$\text{ Length of canvass}=\frac{14220\text{ cm}}{100}\times\frac{m}{cm}$

$\text{ Length of canvass}=142.20\text{ m}$

Therefore, 142.2 meters of canvas of width 80 cm required to cover 150 such boxes.

5 0

2 years ago

Question 6 The mineral content of a particular brand of supplement pills is normally distributed with mean 490 mg and variance o

AysviL [449]

Answer:

0.3085 = 30.85% probability that a randomly selected pill contains at least 500 mg of minerals

Step-by-step explanation:

Normal Probability Distribution:

Problems of normal distributions can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the z-score 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 p-value, we get the probability that the value of the measure is greater than X.

Mean 490 mg and variance of 400.

This means that $\mu = 490, \sigma = \sqrt{400} = 20$

What is the probability that a randomly selected pill contains at least 500 mg of minerals?

This is 1 subtracted by the p-value of Z when X = 500. So

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

$Z = \frac{500 - 490}{20}$

$Z = 0.5$

$Z = 0.5$ has a p-value of 0.6915.

1 - 0.6915 = 0.3085

0.3085 = 30.85% probability that a randomly selected pill contains at least 500 mg of minerals

4 0

1 year ago

4. Hydrocarbons in the cab of an automobile were measured during trips on the New Jersey Turnpike and trips through the Lincoln

pochemuha

Answer:

No these these result do not differ at 95% confidence level

Step-by-step explanation:

From the question we are told that

The first concentrations is $c _1= 30.0 \ g/m^3$

The second concentrations is $c _2 = 52.9 \ g/m^3$

The first sample size is $n_1 = 32$

The second sample size is $n_2 = 32$

The first standard deviation is $\sigma_1 = 30.0$

The first standard deviation is $\sigma_1 = 29.0$

The mean for Turnpike is $\= x _1 = \frac{c_1}{n} = \frac{31.4}{32} = 0.98125$

The mean for Tunnel is $\= x _2 = \frac{c_2}{n} = \frac{52.9}{32} = 1.6531$

The null hypothesis is $H_o : \mu _1 - \mu_2 = 0$

The alternative hypothesis is $H_a : \mu _1 - \mu_2 \ne 0$

Generally the test statistics is mathematically represented as

$t = \frac{\= x_1 - \= x_2}{ \sqrt{\frac{\sigma_1^2}{n_1} +\frac{\sigma_2^2}{n_2} }}$

$t = \frac{0.98125 - 1.6531}{ \sqrt{\frac{30^2}{32} +\frac{29^2}{32} }}$

$t = - 0.0899$

Generally the degree of freedom is mathematically represented as

$df = 32+ 32 - 2$

$df = 62$

The significance $\alpha$ is evaluated as

$\alpha = (C - 100 )\%$

=> $\alpha = (95 - 100 )\%$

=> $\alpha =0.05$

The critical value is evaluated as

$t_c = 2 * t_{0.05 , 62}$

From the student t- distribution table

$t_{0.05, 62} = 1.67$

So

$t_c = 2 * 1.67$

=> $t_c = 3.34$

given that

$t_c > t$ we fail to reject the null hypothesis so this mean that the result do not differ

6 0

1 year ago

15,000 blue trout were released into the Meherrin River for a scientific study. The function, f(x) = 15,000 (98)x , represents t

Norma-Jean [14]

Answer:

I'm guessing you mean f(x)=15,000(9/8)^x. If this is what you mean, the population would increase by about 12,000 (12030.4870605 to be exact).

Step-by-step explanation:

Starting equation: f(x)=15,000(9/8)^x

You can clean up the 9/8 to be 1.125

Now what you want to do is find the answer to (9/8)^5 which is 1.8020324707

Next multiply 1.8020324707 by 15,000 and you get 27030.4870605

Finally 27,030.4870605 - 15,000 gives you 12030.4870605. Which means that the population increased by about 12,000.

0 0

2 years ago