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3 years ago

14

a recipe requires 3.5 teaspoons of sugar to make a tart which equation shows the number of teaspoons of sugar , y , needed to ma

ke x tarts?

1 answer:

Natalka [10]3 years ago

6 0

Answer: x = 3.5y

Step-by-step explanation:

3.5 teaspoons of sugar (y) per tart (x) would look like this

You multiply 3.5 by y to equal x divided by 3.5

x = 3.5y

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A model length of 12 cm. represents an actual length of 102 ft.the model has a width of 48cm. What is the width of the actual ob

Elodia [21]

the answer would be 408

just set it up like this

12 48

---- = ----

102 ?

12*4=48 so 102*4 = 408

8 0

4 years ago

Jen picked 3/4 of a gallon of strawberries in half an hour.If she keeps picking strawberries at the same time, how many gallons

MrRa [10]

Answer:

Number of gallons of strawberries will she have picked in 2 hours is, 3

Step-by-step explanation:

Unit rate defined as the rate are expressed as a quantity of 1, such as 4 feet per second or 6 miles per hour, they are called unit rates.

Given the statement: Jen picked 3/4 of a gallon of strawberries in half an hour.

⇒ In $\frac{1}{2}$ hour Jen picked $\frac{3}{4}$ gallon of strawberries.

Unit rate per hour = $\frac{\frac{3}{4} }{\frac{1}{2}} =\frac{3}{4} \times \frac{2}{1} = \frac{6}{4} = \frac{3}{2}$

To find the number of gallons will she have picked in 2 hours.

$Number of gallons = unit rate per hours \times 2$

= $\frac{3}{2} \times 2 = 3$

Therefore, the number of gallons will she have picked in 2 hours is, 3

6 0

3 years ago

The amount of coffee that a filling machine puts into an 8-ounce jar is normally distributed with a mean of 8.2 ounces and a sta

nordsb [41]

Answer:

73.30% probability that the sampling error made in estimating the mean amount of coffee for all 8-ounce jars by the mean of a random sample of 100 jars will be at most 0.02 ounce

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal probability distribution

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.

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean $\mu$ and standard deviation $\sigma$ , the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$ .

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

In this problem, we have that:

$\mu = 8.2, \sigma = 0.18, n = 100, s = \frac{0.18}{\sqrt{100}} = 0.018$

What is the probability that the sampling error made in estimating the mean amount of coffee for all 8-ounce jars by the mean of a random sample of 100 jars will be at most 0.02 ounce?

This is the pvalue of Z when X = 8.2 + 0.02 = 8.22 subtracted by the pvalue of Z when X = 8.2 - 0.02 = 8.18. So

X = 8.22

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

By the Central Limit Theorem

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

$Z = \frac{8.22 - 8.2}{0.018}$

$Z = 1.11$

$Z = 1.11$ has a pvalue of 0.8665

X = 8.18

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

$Z = \frac{8.18 - 8.2}{0.018}$

$Z = -1.11$

$Z = -1.11$ has a pvalue of 0.1335

0.8665 - 0.1335 = 0.7330

73.30% probability that the sampling error made in estimating the mean amount of coffee for all 8-ounce jars by the mean of a random sample of 100 jars will be at most 0.02 ounce

8 0

3 years ago

Question 8 (1 point)

ra1l [238]

Probly the first one it looks right

6 0

3 years ago

The state education commission wants to estimate the fraction of tenth grade students that have reading skills at or below the e

I am Lyosha [343]

Answer:

(a) The proportion of tenth graders reading at or below the eighth grade level is 0.1673.

(b) The 95% confidence interval for the population proportion of tenth graders reading at or below the eighth grade level is (0.198, 0.260).

Step-by-step explanation:

Let <em>X</em> = number of students who read above the eighth grade level.

(a)

A sample of <em>n</em> = 269 students are selected. Of these 269 students, <em>X</em> = 224 students who can read above the eighth grade level.

Compute the proportion of students who can read above the eighth grade level as follows:

$\hat p=\frac{X}{n}=\frac{224}{269}=0.8327$

The proportion of students who can read above the eighth grade level is 0.8327.

Compute the proportion of tenth graders reading at or below the eighth grade level as follows:

$1-\hat p=1-0.8327$

$=0.1673$

Thus, the proportion of tenth graders reading at or below the eighth grade level is 0.1673.

(b)

the information provided is:

<em>n</em> = 709

<em>X</em> = 546

Compute the sample proportion of tenth graders reading at or below the eighth grade level as follows:

$\hat q=1-\hat p$

$=1-\frac{X}{n}$

$=1-\frac{546}{709}$

$=0.2299\\\approx 0.229$

The critical value of <em>z</em> for 95% confidence interval is:

$z_{\alpha/2}=z_{0.05/2}=z_{0.025}=1.96$

Compute the 95% confidence interval for the population proportion as follows:

$CI=\hat p \pm z_{\alpha/2}\sqrt{\frac{\hat p(1-\hat p)}{n}}$

$=0.229\pm 1.96\times \sqrt{\frac{0.229(1-0.229)}{709}}\\=0.229\pm 0.03136\\=(0.19764, 0.26036)\\\approx (0.198, 0.260)$

Thus, the 95% confidence interval for the population proportion of tenth graders reading at or below the eighth grade level is (0.198, 0.260).

5 0

3 years ago