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2 years ago

A supermarket soup can display consists of 5 levels of

1 answer:

8 0

A supermarket shop cans display consists of 5 levels of stacked cans. In the display, each level has 4 more cans than the level above it-the total number of cans in display are (b) 100

Step-by-step explanation:

Given that ,

The display consists of 5 levels of stacked cans- which means that there are 5 levels in which the stacked cans are displayed .

so,the no:of levels are 5

Now given that level has 4 more cans than the level above it.

so. the number of cans in a single level = 5*4=20

Now we know that the total number of level is 5 so we multiply 20 withe the number of levels of can stacked

==>20*5=100

Thus we can say that the total number of cans in display are (b) 100

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Sid need 0.6 meters of canvas material to make a carry-all bag for his wheelchair.if canvas is $10.09 per meter, how much will s

Olegator [25]

$6.05 because 10.09 × 0.6 is 6.054, round down because the last number is less than 5, and you have 6.05.

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2 years ago

A line passes through the points (15,−13) and (16,−11). Hollis writes the equation y+13=(x−15) to represent the line. Which answ

Andru [333]

The equation formula is y - y1 = m(x-x1)

Using the first point for x1, y1:

Y +13 = m(x-15)

M is the slope which is the change in y over the change in x:

M = -11–13 / 16-15 = 2/1 = 2

The equation becomes y +13 =2(x+15)

The answer is:

He incorrectly wrote the slope in his equation. He should have written y+13=2(x−15).

8 0

2 years ago

The temperature reading from a thermocouple placed in a constant-temperature medium is normally distributed with mean μ, the act

stealth61 [152]

Answer: 0.2551

Step-by-step explanation:

Given : The temperature reading from a thermocouple placed in a constant-temperature medium is normally distributed with mean μ, the actual temperature of the medium, and standard deviation σ.

Significance level : $\alpha=1-0.95=0.05$

The critical z-value for 95% confidence : $z_{\alpha/2}=1.960$ (1)

Since , $z=\dfrac{x-\mu}{\sigma}$ (where x be any random variable that represents the temperature reading from a thermocouple.)

Then, from (1)

$\dfrac{x-\mu}{\sigma}=1.96\\\\ x-\mu=1.96\sigma$ (2)

Also, all readings are within 0.5° of μ,

i.e. $x-\mu$

i.e. $1.96\sigma$ [From (2)]

i.e. $\sigma$

i.e. $\sigma\approx0.2551$

The required standard deviation : $\sigma=0.2551$

3 0

2 years ago

Matt wants to buy a television. If he pays cash, he gets a discount of 7%. If he pays with a loan he has to pay an extra 10% in

Vitek1552 [10]

Answer:

Cost of Television = $294

Step-by-step explanation:

7% discount means 7/100 = 0.07 LESS

So we need to multiply by 1 - 0.07 = 0.93

Also, 10% MORE, or extra, means 10/100 = 0.10 MORE

So we need to multiply by 1 + 0.10 = 1.1

Let cost of television be "c", so we can say:

7% discount would mean cost to be c(0.93) = 0.93c
10% with interest would mean cost to be c(1.10) = 1.10c

The difference is 49.98. So we subtract lower value (0.93c) from larger value (1.10c) and equate to 49.98 and solve for c:

$1.10c-0.93c=49.98\\0.17c=49.98\\c=\frac{49.98}{0.17}\\c=294$

Cost of Television = $294

6 0

2 years ago

A study of immunizations among school‑age children in California found that some areas had rates of unvaccinated school‑age chil

Rom4ik [11]

Answer:

Probability that none of the 20 children in such a classroom would be unvaccinated is 0.055.

Step-by-step explanation:

We are given that a classroom of 20 children in one such area where 13.5% of children are unvaccinated.

If there are no siblings in the classroom, we are willing to consider the vaccination status of the 2020 unrelated children to be independent.

The above situation can be represented through binomial distribution;

$P(X=r) = \binom{n}{r} \times p^{r} \times (1-p)^{n-r} ; x = 0,1,2,3,......$

where, n = number of trials (samples) taken = 20 children

r = number of success = none of the 20 children

p = probability of success which in our case is probability that

children are unvaccinated, i.e; p = 13.5%

Let X = Number of children that are unvaccinated

So, X ~ Binom(n = 20, p = 0.135)

Now, Probability that none of the 20 children in such a classroom would be unvaccinated is given by = P(X = 0)

P(X = 0) = $\binom{20}{0} \times 0.135^{0} \times (1-0.135)^{20-0}$

= $1\times 1 \times 0.865^{20}$

= 0.055

Hence, the probability that none of the 20 children in such a classroom would be unvaccinated is 0.055.

8 0

2 years ago