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Marina CMI [18]

3 years ago

10

An ice cream stand sells chocolate, vanilla, and strawberry ice cream as well as a choice of 22 toppings. how many choices are t

here for a single flavor of ice cream with one topping?
A. 25
B.46
C.52
D.66

1 answer:

serg [7]3 years ago

3 0

They sell 3 different flavors of ice cream and 22 toppings.

To find the number of choices, multiply the number of flavors and number of toppings.

3 flavors x 22 toppings = 66 different choices.

The correct answer would be D. 66

You might be interested in

Find the equation of the line parallel to y=2x+6 passing through (4,5)

ozzi

Answer:

y=2x-3

when two lines are parallel, they have the same slope.

in the formula y=mx+b, m=slope of the line

we know 2 will be the slope for the line passing through (4,5)

next plug in (4,5) into the formula with 2 as the slope

4=x 5=y

next 5=2(4)+b

and you end up with

y=2x-3 when you simplify

3 0

2 years ago

If n(A)=26, n(B)=41, and n(A ∩ B)=8, find n(A ∪ B).

miskamm [114]

n (AUB) = 59

Step-by-step explanation:

n(A U B) = n(A) + n(B) -n(A intersection B)

= 26+ 41- 8

= 59

8 0

2 years ago

Val rented a bicycle while she was on vacation she paid a flat rental fee of $55.00 plus $8.50 EACH day the total cost was $123

Tom [10]

Answer:

14 days approximately or 14.47 to be exact

Step-by-step explanation:

Let the number of days be x

Val paid $123 for x days for renting a bicycle where as $8.50 was rent of bicycle per day so dividing $123/8.50 gives us number of days

which are 14.47 days.

$55 amount is ambiguous that whether it was flat ( as in apartment) rental fee or what?

3 0

3 years ago

In 2013, a study found that U.S. consumers consumed soft drinks with a mean of 42.2 gallons per year and variance of 169. Ninety

melisa1 [442]

Answer:

Ninety-five percent of consumers in the U.S. consumed less than 63.59 gallons.

Step-by-step explanation:

Problems of normally distributed samples are 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.

In this problem, we have that:

The standard deviation is the square root of the variance, so $\sigma = \sqrt{169} = 13$

Also, the mean is 42.2, so $\mu = 42.2$

Ninety-five percent of consumers in the U.S. consumed less than how many gallons?

The 95th percentile, which is the value of X when Z has a pvalue of 0.95. So X when $Z = 1.645$

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

$1.645 = \frac{X - 42.2}{13}$

$X - 42.2 = 13*1.645$

$X = 63.59$

Ninety-five percent of consumers in the U.S. consumed less than 63.59 gallons.

4 0

3 years ago

Two machines are used for filling glass bottles with a soft-drink beverage. The filling process have known standard deviations s

stellarik [79]

Answer:

a. We reject the null hypothesis at the significance level of 0.05

b. The p-value is zero for practical applications

c. (-0.0225, -0.0375)

Step-by-step explanation:

Let the bottles from machine 1 be the first population and the bottles from machine 2 be the second population.

Then we have $n_{1} = 25$ , $\bar{x}_{1} = 2.04$ , $\sigma_{1} = 0.010$ and $n_{2} = 20$ , $\bar{x}_{2} = 2.07$ , $\sigma_{2} = 0.015$ . The pooled estimate is given by

$\sigma_{p}^{2} = \frac{(n_{1}-1)\sigma_{1}^{2}+(n_{2}-1)\sigma_{2}^{2}}{n_{1}+n_{2}-2} = \frac{(25-1)(0.010)^{2}+(20-1)(0.015)^{2}}{25+20-2} = 0.0001552$

a. We want to test $H_{0}: \mu_{1}-\mu_{2} = 0$ vs $H_{1}: \mu_{1}-\mu_{2} \neq 0$ (two-tailed alternative).

The test statistic is $T = \frac{\bar{x}_{1} - \bar{x}_{2}-0}{S_{p}\sqrt{1/n_{1}+1/n_{2}}}$ and the observed value is $t_{0} = \frac{2.04 - 2.07}{(0.01246)(0.3)} = -8.0257$ . T has a Student's t distribution with 20 + 25 - 2 = 43 df.

The rejection region is given by RR = {t | t < -2.0167 or t > 2.0167} where -2.0167 and 2.0167 are the 2.5th and 97.5th quantiles of the Student's t distribution with 43 df respectively. Because the observed value $t_{0}$ falls inside RR, we reject the null hypothesis at the significance level of 0.05

b. The p-value for this test is given by $2P(T$ 0 (4.359564e-10) because we have a two-tailed alternative. Here T has a t distribution with 43 df.

c. The 95% confidence interval for the true mean difference is given by (if the samples are independent)

$(\bar{x}_{1}-\bar{x}_{2})\pm t_{0.05/2}s_{p}\sqrt{\frac{1}{25}+\frac{1}{20}}$ , i.e.,

$-0.03\pm t_{0.025}0.012459\sqrt{\frac{1}{25}+\frac{1}{20}}$

where $t_{0.025}$ is the 2.5th quantile of the t distribution with (25+20-2) = 43 degrees of freedom. So

$-0.03\pm(2.0167)(0.012459)(0.3)$ , i.e.,

(-0.0225, -0.0375)

8 0

2 years ago